NCERT Solutions for Class 12 Maths In Hindi Chapter 2 Inverse Trigonometric Functions In Hindi

प्रश्नावली 2.1

निम्नलिखित के मुख्य मानों को ज्ञात कीजिए 

1. \[\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}\]

उत्तर: मान लीजिए कि, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  y}}$

अतः ${\text{sin y  =   -  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{  =   -  sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{  =  sin }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ होता है और ${\text{sin }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{6}}}} \right){\text{  =   -  }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |इसलिए, \[{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right)\]  का प्रमुख मान \[{\text{ -  }}\dfrac{\pi}{{\text{6}}}\] हैं

2. \[{\mathbf{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {\text{3}} }}{{\text{2}}}} \right)}\]

उत्तर:  मान लीजिए कि, ${\cos ^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt 3 }}{{\text{2}}}} \right){\text{  =  y}}$

अतः ${\text{sin y  = }}\dfrac{{\sqrt {\text{3}} }}{{\text{2}}}{\text{  =  cos  }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{sin }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{6}}}} \right){\text{  =   -  }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, \[{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {\text{3}} }}{{\text{2}}}} \right)\]  का प्रमुख मान \[\dfrac{\pi}{{\text{6}}}\] हैं

3. \[\mathbf{{\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( 2 \right)}\] 

उत्तर:  मान लीजिए कि, ${\text{cosec}}{{\text{ }}^{{\text{ - 1}}}}\left( {\text{2}} \right){\text{  =  y}}$

अतः ${\text{cosec y  =  2   =  cosec  }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$हमें ज्ञात है कि ${\text{cose}}{{\text{c}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]{\text{  -  }}\left\{ {\text{0}} \right\}$ होता है और ${\text{cosec }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{  =  2 }}$ हैं|

इसलिए, \[{\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{2 }}} \right)\]  का प्रमुख मान \[\dfrac{\pi}{{\text{6}}}\] हैं

4. \[\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\sqrt {\text{3}} } \right)}\]

उत्तर:  मान लीजिए कि, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\sqrt {\text{3}} } \right){\text{  =  y}}$

अतः ${\text{tan y  =   -  }}\sqrt {\text{3}} {\text{  =   -  tan }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{  =  tan }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ होता है और ${\text{tan }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{3}}}} \right){\text{  =   -  }}\sqrt {\text{3}} $ हैं

इसलिए, \[{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\sqrt {\text{3}} } \right)\]  का प्रमुख मान \[{\text{ -  }}\dfrac{\pi}{3}\] हैं

5. \[\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}\] 

उत्तर:  मान लीजिए कि, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  y}}$

अतः ${\text{cos y  =   -  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{  =   -  cos }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{  =  cos }}\left( {{{\pi  - }}\dfrac{\pi}{{\text{3}}}} \right){\text{  =  cos }}\left( {\dfrac{{{{2 \pi }}}}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{cos }}\left( {{\text{  -  }}\dfrac{{{{2 \pi }}}}{{\text{3}}}} \right){\text{  =   -  }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, \[{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right)\]  का प्रमुख मान \[\dfrac{{{{2\pi }}}}{3}\] हैं

6. \[\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ -  1 }}} \right)}\]

उत्तर:  मान लीजिए कि, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ -  1}}} \right){\text{  =  y}}$

अतः ${\text{tan y  =   -  1  =   -  tan }}\left( {\dfrac{\pi}{4}} \right){\text{  =  tan }}\left( {{\text{  -  }}\dfrac{\pi}{4}} \right)$

 हमें ज्ञात है कि ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ होता है और ${\text{tan }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{4}}}} \right){\text{  =   -  1 }}$ हैं |

 इसलिए, \[{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ -  1 }}} \right)\]  का प्रमुख मान \[{\text{ -  }}\dfrac{\pi}{4}\] हैं

7. \[\mathbf{{\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}} \right)}\]

उत्तर:  मान लीजिए कि, ${\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}} \right){\text{  =  y}}$

अतः ${\text{sec y  =  }}\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}{\text{  =  sec }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{se}}{{\text{c}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]{\text{  -  }}\left\{ {\dfrac{\pi}{{\text{2}}}} \right\}$ होता है और ${\text{sec }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{  =  }}\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}$ हैं 

इसलिए, \[{\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}} \right)\]  का प्रमुख मान \[\dfrac{\pi}{{\text{6}}}\] हैं

8. \[\mathbf{{\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right)}\] 

उत्तर:  मान लीजिए कि, ${\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{  =  y}}$

अतः ${\text{sin y  =  }}\sqrt {\text{3}} {\text{  =  cot }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{t}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{cot }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{  =  }}\sqrt {\text{3}} $ हैं |

इसलिए, \[{\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right)\]  का प्रमुख मान \[\dfrac{\pi}{{\text{6}}}\] हैं

9. \[\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}} \right)}\] 

उत्तर:  मान लीजिए कि, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}} \right){\text{  =  y}}$

अतः ${\text{cos y  =   -  }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}{\text{  =   -  cos }}\left( {\dfrac{\pi}{{\text{4}}}} \right){\text{ =  cos }}\left( {{{\pi  - }}\dfrac{\pi}{{\text{4}}}} \right){\text{  =  cos }}\left( {\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{{0 , \pi }}} \right]$ होता है और ${\text{cos  }}\left( {\dfrac{{{{3\pi }}}}{{\text{4}}}} \right){\text{  =   -  }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}$ हैं |

इसलिए, \[{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}} \right)\]  का प्रमुख मान \[\dfrac{{{{3\pi }}}}{{\text{4}}}\] हैं

10. \[\mathbf{{\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\sqrt {\text{2}} } \right)}\] 

उत्तर: मान लीजिए कि, ${\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\sqrt {\text{2}} } \right){\text{  =  y}}$

अतः ${\text{cosec y  =   -  }}\sqrt {\text{2}} {\text{  =   -  cosec }}\left( {\dfrac{\pi}{{\text{4}}}} \right){\text{  =  cosec }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{4}}}} \right)$

हमें ज्ञात है कि ${\text{cose}}{{\text{c}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]{\text{  -  }}\left\{ {\text{0}} \right\}$ होता है और ${\text{cosec }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{4}}}} \right){\text{  =   -  }}\sqrt {\text{2}} $ हैं |

इसलिए, \[{\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\sqrt {\text{2}} } \right)\]  का प्रमुख मान \[\dfrac{\pi}{4}\] हैं

निम्नलिखित के मुख्य मानों को ज्ञात कीजिए

11. $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{  +  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  +  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}$

 उत्तर:  मान लीजिए कि, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{  =  x  }}$

 अतः ${\text{tan x  =  1  =  tan }}\left( {\dfrac{\pi}{{\text{4}}}} \right)$

 हमें ज्ञात है कि ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left( {{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right)$ होता है और ${\text{tan }}\left( {\dfrac{\pi}{{\text{4}}}} \right){\text{  =  1 }}$ हैं |

इसलिए, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{  =  }}\dfrac{\pi}{{\text{4}}}{\text{ }}$

मान लीजिए कि, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  y }}$

अतः ${\text{cos y  =   -  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{  =   -  cos }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{  =   -  cos }}\left( {{{\pi   -  }}\dfrac{\pi}{{\text{3}}}} \right){\text{  =  cos }}\left( {\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{cos }}\left( {\dfrac{{{{2\pi }}}}{{\text{3}}}} \right){\text{  =   -  }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  }}\dfrac{{{{2\pi }}}}{{\text{3}}}{\text{ }}$

मान लीजिए कि, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  z }}$

अतः ${\text{sin z   =   -  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{  =   -  sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{  =  sin }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left( {{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right)$ होता है और ${\text{sin }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{6}}}} \right){\text{  =   -  }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =   -  }}\dfrac{\pi}{{\text{6}}}{\text{ }}$

अब, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{  +  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  +  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right)$ ${\text{ =  }}\dfrac{\pi}{{\text{4}}}{\text{  +  }}\dfrac{{{{2\pi }}}}{{\text{3}}}{\text{  -  }}\dfrac{\pi}{{\text{6}}}$

${\text{ =  }}\dfrac{{{{3\pi   +  8\pi   -  2\pi }}}}{{{\text{12}}}}$ 

${\text{ =  }}\dfrac{{{{9\pi}}}}{{{\text{12}}}}{\text{  =  }}\dfrac{{{{3\pi}}}}{{\text{4}}}$

12. $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  +  2si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)}$

उत्तर:  मान लीजिए कि, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  x }}$

अतः ${\text{cos x   =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{  =  cos }}\left( {\dfrac{\pi}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{cos }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{  =  }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं 

इसलिए, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  }}\dfrac{\pi}{{\text{3}}}{\text{ }}$

मान लीजिए कि, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  y }}$

अतः ${\text{sin y   =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{  =  sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left( {{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right)$ होता है और ${\text{sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{  =  }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  }}\dfrac{\pi}{{\text{6}}}{\text{ }}$

अब, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  +  2si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)$ ${\text{ =  }}\dfrac{\pi}{{\text{3}}}{\text{  +  2  X  }}\dfrac{\pi}{{\text{6}}}$

    ${\text{ =  }}\dfrac{\pi}{{\text{3}}}{\text{  +  }}\dfrac{\pi}{{\text{3}}}{\text{  =  }}\dfrac{{{{2\pi }}}}{{\text{3}}}$  

 13.  यदि $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  =  y}}}$, तो 

1. $\mathbf{{\text{0}} \leqslant {\text{y}} \leqslant \pi}$

2. $ \mathbf{- \;\dfrac{\pi}{{\text{2}}} \leqslant {\text{y}} \leqslant \dfrac{\pi}{{\text{2}}}}$

3. $\mathbf{{{0  <  y  <  \pi }}}$

4. $\mathbf{{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{  <  y  <  }}\dfrac{\pi}{{\text{2}}}}$

उत्तर:  हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left( {{\text{ -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right)$ होता है

इसलिए, $ - \;\dfrac{\pi}{{\text{2}}} \leqslant {\text{y}} \leqslant \dfrac{\pi}{{\text{2}}}$ 

अतः विकल्प (B ) सही उत्तर हैं |

14. $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{  -  se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{  -  2}}} \right)}$ का मान बराबर है 

1. $\mathbf{\pi}$

2. $\mathbf{{\text{ - }}\;\dfrac{\pi}{{\text{3}}}}$

3. $\mathbf{\dfrac{\pi}{3}}$

4. $\mathbf{\dfrac{{{{2\pi }}}}{3}}$

उत्तर:  दिया गया है,  ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{  -  se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{  -  2}}} \right)$

मान लीजिए कि, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{  =  x }}$

अतः ${\text{tan x   =  }}\sqrt {\text{3}} {\text{  =  tan }}\left( {\dfrac{\pi}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{  -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ होता है और ${\text{tan }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{  =  }}\sqrt {\text{3}} $ हैं |

इसलिए, \[{\tan ^{{\text{ - 1}}}}\left( {\sqrt 3 } \right){\text{  =  }}\dfrac{\pi}{{\text{3}}}{\text{ }}\]

मान लीजिए कि, ${\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{  -  2}}} \right){\text{  =  y }}$

अतः ${\text{sin y   =   -  2  =   -  sec }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{  =  sec }}\left( {{{\pi   -  }}\dfrac{\pi}{{\text{3}}}} \right){\text{  =  sec }}\left( {\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{sec}}{{\text{ }}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]{\text{  -  }}\left\{ {\dfrac{\pi}{{\text{2}}}} \right\}$ होता है और ${\text{sec }}\left( {\dfrac{{{{2\pi }}}}{{\text{3}}}} \right){\text{  =   -  2 }}$ हैं 

इसलिए, ${\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{  -  2}}} \right){\text{  =  }}\dfrac{{{{2\pi }}}}{{\text{3}}}{\text{ }}$

अब, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{  -  se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{  -  2}}} \right)$${\text{ =  }}\dfrac{\pi}{{\text{3}}}{\text{  -  }}\dfrac{{{{2\pi }}}}{{\text{3}}}{\text{  =   -  }}\dfrac{\pi}{{\text{3}}}$

अत: विकल्प (B) सही हैं | 

प्रश्नावली 2.2

निम्नलिखित को सिद्ध कीजिए 

1. $\mathbf{{\text{3si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  =  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{3x  -  4}}{{\text{x}}^{\text{3}}}} \right){\text{, x}}\; \in \;\;\left[ {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{,}}\dfrac{{\text{1}}}{{\text{2}}}} \right]}$

उत्तर: ${{x  =  sin \theta }}$ 

$ \Rightarrow {\text{ si}}{{\text{n}}^{{\text{ - 1}}}}{{x  =  \theta }}$

${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{3x  -  4}}{{\text{x}}^{\text{3}}}} \right){\text{  =  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{3sin x  -  4si}}{{\text{n}}^{\text{3}}}{\text{ x}}} \right)$

$ \Rightarrow {\text{ si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{{sin 3\theta }}} \right)$

$ \Rightarrow {\text{ 3}}\theta $

$ \Rightarrow {\text{ 3 si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

2. $\mathbf{{\text{3co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x  =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{4}}{{\text{x}}^{\text{3}}}{\text{ - 3x}}} \right){\text{, x}}\; \in \;\;\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{, 1 }}} \right]}$

उत्तर: ${{x  =  cos }}\theta$  

$ \Rightarrow {\text{ co}}{{\text{s}}^{{\text{ - 1}}}}{{x  =  \theta }}$

${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{4}}{{\text{x}}^{\text{3}}}{\text{  -  3x}}} \right){\text{  =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{4 co}}{{\text{s}}^{\text{3}}}{{ \theta   -  3 cos \theta }}} \right)$

 $ \Rightarrow {\text{ co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{{cos 3\theta }}} \right)$

$ \Rightarrow {\text{ 3}}\theta $

$ \Rightarrow {\text{ 3 co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}$

3. $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{2}}}{{{\text{11}}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{7}}}{{{\text{24}}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}}$ 

उत्तर: $ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{2}}}{{{\text{11}}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{7}}}{{{\text{24}}}}$

$ \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{{\text{2}}}{{{\text{11}}}}{\text{  +  }}\dfrac{{\text{7}}}{{{\text{24}}}}}}{{{\text{1  -  }}\dfrac{{\text{2}}}{{{\text{11}}}}{\text{ }}{\text{. }}\dfrac{{\text{7}}}{{{\text{24}}}}}}$ 

$ \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{{{\text{48  +  77}}}}{{{{11  \times  24 }}}}}}{{\dfrac{{{{11  \times  24  -  14}}}}{{{{11  \times24 }}}}}}$ 

$ \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{48  +  77}}}}{{{\text{264  -  14}}}}$${\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{125}}}}{{{\text{250 }}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}$

4: $\mathbf{{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{31}}}}{{{\text{17}}}}}$

उत्तर: $ \Rightarrow 2\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{1}{2}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{1}{7}$ 

$ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2}}{\text{.}}\dfrac{{\text{1}}}{{\text{2}}}}}{{{\text{1  -  }}{{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)}^{\text{2}}}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}$

$ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{1}{{\left( {\dfrac{3}{4}} \right)}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}$

$ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{4}{3}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}$

$ \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{4}{3}{\text{  +  }}\dfrac{1}{7}}}{{{\text{1  -  }}\dfrac{4}{3}{\text{ }}{\text{. }}\dfrac{1}{7}}}$ 

$ \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{{{\text{28  +  3}}}}{{{\text{21 }}}}}}{{\dfrac{{{\text{21  -  4}}}}{{{\text{21 }}}}}}$ 

$ \Rightarrow {\text{ ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{31}}}}{{17}}$

निम्नलिखित फलनों को सरलतम रूप में लिखिए 

5: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1  +  }}{{\text{x}}^{\text{2}}}} {\text{  -  1}}}}{{\text{x}}}{\text{, x}} \ne {\text{0 }}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1  +  }}{{\text{x}}^{\text{2}}}} {\text{  -  1}}}}{{\text{x}}}{\text{,x}} \ne {\text{0}}$

${{x  =  tan\theta }} \Rightarrow {{\theta   =  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

$\therefore {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} {\text{ - 1}}}}{{\text{x}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1  +  ta}}{{\text{n}}^{\text{2}}}{{\theta }}} {\text{  -  1}}}}{{{{tan\theta }}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{{(sec \theta   -  1)}}}}{{{{tan\theta }}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{{(1  -  cos\theta )}}}}{{{{sin\theta }}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2si}}{{\text{n}}^{\text{2}}}\dfrac{{{\theta }}}{{\text{2}}}}}{{{\text{2sin}}\dfrac{{{\theta }}}{{\text{2}}}{\text{cos}}\dfrac{{{\theta }}}{{\text{2}}}}}$ 

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{{\theta }}}{{\text{2}}}} \right){\text{  =  }}\dfrac{{{\theta }}}{{\text{2}}}{\text{  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

6: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{  -  1}}} }},\;\left| {{\text{ x }}} \right|{\text{  >  1}}}$

उत्तर: ${{x  =  cosec\theta }} \Rightarrow {{\theta   =  cose}}{{\text{c}}^{{\text{ - 1}}}}{\text{x}}$

$\begin{align} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{  -  1}}} }}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\sqrt {{\text{cose}}{{\text{c}}^{\text{2}}}{{\theta   -  1}}} }} \hfill \\ {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{{{cot\theta }}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}{{(tan\theta )}} \hfill \\ {{ =  \theta   =  cose}}{{\text{c}}^{{\text{ - 1}}}}{\text{x  =  }}\dfrac{\pi}{{\text{2}}}{\text{  -  se}}{{\text{c}}^{{\text{ - 1}}}}{\text{x}} \hfill \\  \end{align} $

7: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\dfrac{{{\text{1  -  cosx}}}}{{{\text{1  +  cosx}}}}} {{ , 0  <  x  <  \pi }}}$

उत्तर: $ \Rightarrow {\text{ ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\dfrac{{{\text{1  -  cosx}}}}{{{\text{1  +  cosx}}}}}  = $${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{2co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}} \right)$

$ \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}}}} \right){\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right)$

$ \Rightarrow \dfrac{{\text{x}}}{{\text{2}}}$

8: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{cos x  -  sin x}}}}{{{\text{cos x  +  sin x}}}}} \right)}$

उत्तर:  $ \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1  -  }}\dfrac{{{\text{sin x}}}}{{{\text{cos x}}}}}}{{{\text{1  +  }}\dfrac{{{\text{sin x}}}}{{{\text{cos x}}}}}}} \right)$ 

$ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1  -  tan x}}}}{{{\text{1  +  tan x}}}}} \right)$

$ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{  -  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan x}}} \right)$

$ \Rightarrow \dfrac{\pi}{{\text{4}}}{\text{  -  x}}$

9: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{  -  }}{{\text{x}}^{\text{2}}}} }}{\text{ , }}\left| {{\text{ x }}} \right|{\text{  <  a}}}$  

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{  -  }}{{\text{x}}^{\text{2}}}} }}{\text{ , }}\left| {{\text{ x }}} \right|{\text{  <  a}}$

$ \Rightarrow {\text{ ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{  -  }}{{\text{x}}^{\text{2}}}} }}$

$ \Rightarrow {{x  =  a sin \theta }}$

$ \Rightarrow \dfrac{{\text{x}}}{{\text{a}}}{{  =  sin \theta   =  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\text{a}}}} \right)$ 

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{  -  }}{{\text{x}}^{\text{2}}}} }}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{{a sin \theta }}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ -  }}{{\text{a}}^{\text{2}}}{\text{si}}{{\text{n}}^{\text{2}}}{{\theta }}} }}} \right)$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{{a sin \theta }}}}{{{\text{a}}\sqrt {{\text{1  -  si}}{{\text{n}}^{\text{2}}}{{\theta }}} }}} \right){\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{{a sin \theta }}}}{{{{a cos \theta }}}}$ 

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{(tan \theta )  =  \theta   =  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{a}}} $

10:$\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{2}}}{\text{x  -  }}{{\text{x}}^{\text{3}}}}}{{{{\text{a}}^{\text{3}}}{\text{  -  3a}}{{\text{x}}^{\text{2}}}}}} \right){\text{,  a  >  0 ;   -  }}\dfrac{{\text{a}}}{{\sqrt {\text{3}} }}{\text{  <  x  <  }}\dfrac{{\text{a}}}{{\sqrt {\text{3}} }}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{2}}}{\text{x  -  }}{{\text{x}}^{\text{3}}}}}{{{{\text{a}}^{\text{3}}}{\text{  -  3a}}{{\text{x}}^{\text{2}}}}}} \right)$

$ \Rightarrow {{x  =  a tan \theta }}$

 $\Rightarrow \dfrac{{\text{x}}}{{\text{a}}}{{  =  tan \theta }} \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{a}}}$

$ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{2}}}{\text{x  -  }}{{\text{x}}^{\text{3}}}}}{{{{\text{a}}^{\text{3}}}{\text{  -  3a}}{{\text{x}}^{\text{2}}}}}} \right){\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{2}}}{{.a tan \theta   -  }}{{\text{a}}^{\text{3}}}{\text{ta}}{{\text{n}}^{\text{3}}}{{\theta }}}}{{{{\text{a}}^{\text{3}}}{\text{  -  3a}}{\text{.}}{{\text{a}}^{\text{2}}}{\text{ta}}{{\text{n}}^{\text{2}}}{{\theta }}}}} \right)$

$ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{3}}}{{tan \theta   -  }}{{\text{a}}^{\text{3}}}{\text{ta}}{{\text{n}}^{\text{3}}}{{\theta }}}}{{{{\text{a}}^{\text{3}}}{\text{  -  3}}{{\text{a}}^{\text{3}}}{\text{ta}}{{\text{n}}^{\text{2}}}{{\theta }}}}} \right)$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{{3 tan \theta   -  ta}}{{\text{n}}^{\text{3}}}{{\theta }}}}{{{\text{1  -  3ta}}{{\text{n}}^{\text{2}}}{{\theta }}}}} \right)$

$ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{(tan 3\theta )}}$

$\Rightarrow {{3\theta }}$

$ \Rightarrow {\text{3 ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{a}}} $

निम्नलिखित में से प्रत्येक का मान ज्ञात कीजिए

11:\[\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {{\text{2 si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right]}\]

उत्तर:  \[{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {{\text{2 si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right]\]

$\Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ =  x}}$

$ \Rightarrow {\text{sin x  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{  =  sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

$ \Rightarrow \therefore {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{\pi}{{\text{6}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {{\text{2 si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right]{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {{\text{2 }}\left( {{\text{x}}\dfrac{\pi}{{\text{6}}}} \right)} \right)} \right]$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {\dfrac{\pi}{{\text{3}}}} \right)} \right]{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 x}}\dfrac{{\text{1}}}{{\text{2}}}} \right]$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{1  =  }}\dfrac{\pi}{{\text{4}}}$

12: \[\mathbf{{\text{cot}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{a  +  co}}{{\text{t}}^{{\text{ - 1}}}}{\text{a}}} \right)}\]

उत्तर:  \[{\text{cot}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{a  +  co}}{{\text{t}}^{{\text{ - 1}}}}{\text{a}}} \right)\]

$\cot\frac{\pi}{2}   \;\;\;\;[\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2} ]=0$

13: \[\mathbf{{\text{tan}}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1  +  }}{{\text{x}}^{\text{2}}}}}{\text{  +  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1  -  }}{{\text{y}}^{\text{2}}}}}{{{\text{1  +  }}{{\text{y}}^{\text{2}}}}}} \right]{\text{ ,    }}\left| {{\text{ x }}} \right|{\text{  <  1 , y  > 0 or xy  <  1}}}\]

उत्तर:  \[{\text{tan}}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1  +  }}{{\text{x}}^{\text{2}}}}}{\text{  +  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1  -  }}{{\text{y}}^{\text{2}}}}}{{{\text{1  +  }}{{\text{y}}^{\text{2}}}}}} \right]\]

${{x  =  tan \theta  }}$

$ \Rightarrow {{\theta   =  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1  +  }}{{\text{x}}^{\text{2}}}}}{\text{  =  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{{2 tan \theta }}}}{{{\text{1  +  ta}}{{\text{n}}^{\text{2}}}\theta }}} \right){\text{  =  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{{sin 2\theta }}} \right)$

$ {{      =  2\theta   =  2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

$ {\text{y  =  tan}}\emptyset$

$  \Rightarrow \emptyset \;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y}}$

$\therefore \;\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1  -  }}{{\text{y}}^{\text{2}}}}}{{{\text{1  +  }}{{\text{y}}^{\text{2}}}}}{\text{  =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1  -  tan}}{\emptyset ^{\text{2}}}}}{{{\text{1  +  tan}}{\emptyset ^{\text{2}}}}}} \right){\text{  =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos2}}\emptyset } \right){\text{ = 2}}\emptyset$

$ {\text{     =  2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y}} $

$\therefore \;\;{\text{tan}}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1  +  }}{{\text{x}}^{\text{2}}}}}{\text{  +  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1  -  }}{{\text{y}}^{\text{2}}}}}{{{\text{1  +  }}{{\text{y}}^{\text{2}}}}}} \right]$

$ {\text{ =  tan}}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{2 ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  +  2 ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y}}} \right]$

$ {\text{ =  tan}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  +  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y}}} \right]$

$ {\text{ =  tan}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x  +  y}}}}{{{\text{1  -  xy}}}}} \right)} \right]$

  $=\left( {\frac{{{{x  +  y}}}}{{{{1  -  xy}}}}} \right)$ 

14: $\mathbf{{\text{sin }}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{  =  1}}}$

उत्तर: ${\text{sin }}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{  =  1}}$

$\begin{align} \Rightarrow {\text{sin}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}} \right){\text{ cos}}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{  +  cos}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}} \right){\text{ sin}}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right) \hfill \\ {\text{[ sin (A  +  B)  =  sin A cos B  +  cos A sin B ]}} \hfill \\ \dfrac{{\text{1}}}{{\text{5}}}{\text{X x  +  cos}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}} \right){\text{sin}}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{  =  1}} \hfill \\ \dfrac{{\text{x}}}{{\text{5}}}{\text{  +  cos}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}} \right){\text{sin }}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){{  =  1, \ldots  \ldots  \ldots }}{\text{.(1)}} \hfill \\ {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{  =  y}} \hfill \\ {\text{sin y  =  }}\dfrac{{\text{1}}}{{\text{5}}} \Rightarrow {\text{cos y  =  }}\sqrt {{\text{ 1  -  }}{{\left( {\dfrac{{\text{1}}}{{\text{5}}}} \right)}^{\text{2}}}} {\text{  =  }}\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}} \Rightarrow {\text{y  =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}}} \right) \hfill \\ \therefore \;\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{  =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}}} \right){{ \ldots  \ldots  \ldots }}{\text{.(2)}} \hfill \\ {\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x  =  z}} \hfill \\ {\text{cosz  =  x}} \Rightarrow {\text{sinz  =  }}\sqrt {{\text{1  -  }}{{\text{x}}^{\text{2}}}}  \Rightarrow {\text{z  =  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {{\text{1  -  }}{{\text{x}}^{\text{2}}}} } \right) \hfill \\ \therefore \;\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x  =  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {{\text{1  -  }}{{\text{x}}^{\text{2}}}} } \right) \hfill \\  \end{align} $

समीकरण (1),(2) और (3)

$\dfrac{{\text{x}}}{{\text{5}}}{\text{  +  cos }}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}}} \right)} \right){\text{sin }}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right)} \right){\text{  =  1}}$

$\begin{align} \Rightarrow \dfrac{{\text{x}}}{{\text{5}}}{\text{  +  }}\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}}{\text{.}}\sqrt {{\text{1  -  }}{{\text{x}}^{\text{2}}}} {\text{  =  1}} \hfill \\ \Rightarrow {\text{x  +  2}}\sqrt {\text{6}} {\text{.}}\sqrt {{\text{1  -  }}{{\text{x}}^{\text{2}}}} {\text{  =  5}} \hfill \\ \Rightarrow {\text{2}}\sqrt {\text{6}} {\text{.}}\sqrt {{\text{1  -  }}{{\text{x}}^{\text{2}}}} {\text{  =  5  -  x}} \hfill \\ \Rightarrow {\left( {{\text{2}}\sqrt {\text{6}} {\text{.}}\sqrt {{\text{1  -  }}{{\text{x}}^{\text{2}}}} } \right)^{\text{2}}}{\text{  =  }}{\left( {{\text{5  -  x}}} \right)^{\text{2}}} \hfill \\ \Rightarrow \left( {\text{4}} \right)\left( {\text{6}} \right)\;\left( {{\text{1  -  }}{{\text{x}}^{\text{2}}}} \right){\text{  =  25  +  }}{{\text{x}}^{\text{2}}}{\text{  -  10x}} \hfill \\ \Rightarrow {\text{24  -  24}}{{\text{x}}^{\text{2}}}{\text{  =  25  +  }}{{\text{x}}^{\text{2}}}{\text{  -  10x}} \hfill \\ \Rightarrow {\text{25}}{{\text{x}}^{\text{2}}}{\text{  -  10x  +  1  =  0}} \hfill \\ \Rightarrow {\left( {{\text{5x  -  1}}} \right)^{\text{2}}}{\text{  =  0}} \hfill \\ \Rightarrow {\text{5x  -  1  =  0}} \hfill \\ \Rightarrow {\text{5x  =  1}} \hfill \\ \Rightarrow {\text{x  =  }}\dfrac{{\text{1}}}{{\text{5}}} \hfill \\  \end{align} $

15: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1  -  x}}}}{{{\text{x  -  2}}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x  +  1}}}}{{{\text{x  +  2}}}}{\text{  =  }}\dfrac{\pi}{{\text{4}}}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1  -  x}}}}{{{\text{x  -  2}}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x  +  1}}}}{{{\text{x  +  2}}}}{\text{  =  }}\dfrac{\pi}{{\text{4}}}$ 

$\begin{align} \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\dfrac{{\dfrac{{{\text{x  -  1}}}}{{{\text{x  -  2}}}}{\text{  +  }}\dfrac{{{\text{x  +  1}}}}{{{\text{x  +  2}}}}}}{{{\text{1  -  }}\dfrac{{{\text{x  -  1}}}}{{{\text{x  -  2}}}}{\text{.}}\dfrac{{{\text{x  +  1}}}}{{{\text{x  +  2}}}}}}} \right]{\text{  =  }}\dfrac{\pi}{{\text{4}}} \hfill \\ \Rightarrow \left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  +  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x  +  y}}}}{{{\text{1  -  xy}}}}} \right]{\text{  =  }}\dfrac{\pi}{{\text{4}}} \hfill \\ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\dfrac{{\left( {{\text{x  -  1}}} \right)\left( {\;{\text{x  +  2}}} \right)\left( {{\text{x  +  1}}} \right)\left( {{\text{x  -  2}}} \right)}}{{\left( {\;{\text{x  +  2}}} \right)\left( {{\text{x  -  2}}} \right)\left( {{\text{x  -  1}}} \right)\left( {{\text{x  +  1}}} \right)}}} \right]\;{\text{ =  }}\dfrac{\pi}{{\text{4}}} \hfill \\  \end{align} $

$\begin{align} \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\dfrac{{\;{{\text{x}}^{\text{2}}}{\text{  +  x  -  2}}\;{\text{ +  }}{{\text{x}}^{\text{2}}}{\text{  -  x  -  2}}}}{{{{\text{x}}^{\text{2}}}{\text{  -  4  -  }}{{\text{x}}^{\text{2}}}{\text{  +  1}}}}} \right]{\text{  =  }}\dfrac{\pi}{{\text{4}}} \hfill \\ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\dfrac{{{\text{2}}{{\text{x}}^{\text{2}}}{\text{  -  4}}}}{{{\text{ -  3}}}}} \right]{\text{  =  }}\dfrac{\pi}{{\text{4}}} \hfill \\ \end{align} $

$ \Rightarrow {\text{tan}}\left[ {{\text{tan}}\dfrac{{{\text{4  -  2}}{{\text{x}}^{\text{2}}}}}{{\text{3}}}} \right]{\text{  =  tan}}\dfrac{\pi}{{\text{4}}}$

$\begin{align} \Rightarrow \dfrac{{{\text{4  -  2}}{{\text{x}}^{\text{2}}}}}{{\text{3}}}{\text{  =  1}} \hfill \\ \Rightarrow {\text{4  -  2}}{{\text{x}}^{\text{2}}}{\text{  =  3}} \hfill \\ \Rightarrow {\text{2}}{{\text{x}}^{\text{2}}}{\text{  =  4  -  3  =  1}} \hfill \\ \Rightarrow {{x  =   \pm  }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }} \hfill \\  \end{align} $

16 से 18 में प्रत्येक व्यंजक का मान ज्ञात कीजिए 

16: $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)}$

उत्तर:   ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)$ 

$\begin{align} \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin x }}} \right){\text{  =  x ,    x}} \in \left[ {\dfrac{{{{ - \pi }}}}{{\text{2}}},\dfrac{\pi}{{\text{2}}}} \right] \hfill \\ \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right){\text{  =  si}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)} \right]{\text{ = }} \hfill \\ \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}} \right)} \right){\text{ ,          x}} \in \left[ {\dfrac{{{{ - \pi }}}}{{\text{2}}},\dfrac{\pi}{{\text{2}}}} \right] \hfill \\ \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right){\text{  =  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}} \right)} \right)\; = \;\dfrac{\pi}{{\text{3}}} \hfill \\  \end{align} $

17: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)$ 

$\begin{align} \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{{{3\pi }}}}{{\text{4}}}} \right){\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  tan}}\left( {{\text{  -  }}\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)} \right) \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  tan}}\left( {{{\pi   -  }}\dfrac{\pi}{{\text{4}}}} \right)} \right) \hfill \\  \end{align} $

$ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{  -  tan}}\dfrac{\pi}{{\text{4}}}} \right]{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{tan}}\left( {{\text{  -  }}\dfrac{\pi}{{\text{4}}}} \right)} \right]{\text{,     x}} \in \left[ {{\text{  -  }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$

$ \Rightarrow \therefore {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\left( {\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)} \right){\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{tan}}\left( {{\text{  -  }}\dfrac{\pi}{{\text{4}}}} \right)} \right]{\text{  =   -  }}\dfrac{\pi}{{\text{4}}}$

18: $\mathbf{{\text{tan}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  +  co}}{{\text{t}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{2}}}} \right)}$

उत्तर: $ \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  =  x}} \Rightarrow {\text{sin x  =  }}\dfrac{{\text{3}}}{{\text{5}}}$

$\begin{align} \Rightarrow {\text{cos x  =  }}\sqrt {{\text{1  -  si}}{{\text{n}}^{\text{2}}}{\text{x}}} {\text{  =  }}\dfrac{{\text{4}}}{{\text{5}}}\;\; \Rightarrow {\text{sec x  =  }}\dfrac{{\text{5}}}{{\text{4}}} \hfill \\ \Rightarrow \therefore {\text{tan x  =  }}\sqrt {{\text{se}}{{\text{c}}^{\text{2}}}{\text{x  -  1}}} {\text{  =  }}\sqrt {\dfrac{{{\text{25}}}}{{{\text{16}}}}{\text{  -  1}}} {\text{  =  }}\dfrac{{\text{3}}}{{\text{4}}} \hfill \\ \Rightarrow \therefore {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}} \hfill \\ \Rightarrow \therefore {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}}{\text{     }}............\left( {\text{1}} \right) \hfill \\ \Rightarrow {\text{co}}{{\text{t}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{2}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{2}}}{{\text{3}}}{\text{    }}...........\left( {\text{2}} \right) \hfill \\ \Rightarrow {\text{tan}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  +  co}}{{\text{t}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{2}}}} \right) \hfill \\ \Rightarrow {\text{tan}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\left( {\dfrac{{\text{3}}}{{\text{4}}}{\text{  +  }}\dfrac{{\text{2}}}{{\text{3}}}} \right)}}{{{\text{1  -  }}\dfrac{{\text{3}}}{{\text{4}}}{\text{ }}{\text{. }}\dfrac{{\text{2}}}{{\text{3}}}}}} \right)} \right) \hfill \\ \Rightarrow {\text{tan}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{9  +  8}}}}{{{\text{12  -  6}}}}} \right)} \right) \hfill \\ \Rightarrow {\text{tan}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{17}}}}{{\text{6}}}} \right)} \right){\text{  =  }}\dfrac{{{\text{17}}}}{{\text{6}}} \hfill \\  \end{align} $

19: $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)}$ का मान 

1. $\mathbf{\dfrac{{{{7\pi }}}}{{\text{6}}}}$

2. $\mathbf{\dfrac{{{{5\pi }}}}{{\text{6}}}}$

3. $\mathbf{\dfrac{\pi}{3}}$

4. $\mathbf{\dfrac{\pi}{{\text{6}}}}$

उत्तर: ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)\;\;,\;\;\;x \in \left[ {0,\pi } \right]$

$\begin{align} {\text{}} \Rightarrow {\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right){\text{  =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{ -  7\pi }}}}{{\text{6}}}} \right){\text{  =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\left( {{{2\pi   -  }}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)} \right) \hfill \\ {\text{}} \Rightarrow {\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{5 \pi }}}}{{\text{6}}}} \right)\;\;\;\;{\text{,}}\dfrac{{{{5 \pi }}}}{{\text{6}}} \in \left[ {{0, \pi }} \right] \hfill \\ {\text{}} \Rightarrow \therefore \;\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right){\text{  = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{5 \pi }}}}{{\text{6}}}} \right){\text{  =  }}\dfrac{{{{5 \pi }}}}{{\text{6}}} \hfill \\  \end{align} $

 (B) सही विकल्प है  

20: $\mathbf{{\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}{\text{  -  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right)}$ का मान 

1. $\mathbf{\dfrac{1}{2}}$

2. $\mathbf{\dfrac{1}{3}}$

3. $\mathbf{\dfrac{1}{4}}$

4. $\mathbf{1}$

उत्तर: ${\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}{\text{  -  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right)$

$\begin{align} \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =  x}} \hfill \\ \Rightarrow {\text{sin x  =   -  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{  =   -  sin }}\dfrac{\pi}{{\text{6}}}{\text{  =  sin }}\left( {{\text{  -  }}\dfrac{\pi}{{\text{6}}}} \right) \hfill \\ \Rightarrow \therefore \,\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  =   -  }}\dfrac{\pi}{{\text{6}}} \hfill \\ \Rightarrow \therefore \;\;{\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}{\text{  -  si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right){\text{  =  sin}}\left( {\dfrac{\pi}{{\text{3}}}{\text{  +  }}\dfrac{\pi}{{\text{6}}}} \right) \hfill \\ {\text{sin}}\left( {\dfrac{\pi}{{\text{2}}}} \right){\text{  =  1}} \hfill \\ \end{align} $

(D) सही विकल्प है  

21: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{3}} {\text{  -  co}}{{\text{t}}^{{\text{ - 1}}}}{\text{(  -  }}\sqrt {\text{3}} {\text{)}}}$ का मान 

1. \[\mathbf{\pi}\]

2. \[\mathbf{{\text{ -  }}\dfrac{\pi}{{\text{2}}}}\]

3. \[\mathbf{{\text{0}}}\]

4. \[\mathbf{{\text{2}}\sqrt 3} \] 

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{3}} {\text{  -  co}}{{\text{t}}^{{\text{ - 1}}}}{\text{(  -  }}\sqrt {\text{3}} {\text{)}}$

$\begin{align} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{( tan}}\dfrac{\pi}{{\text{3}}}{\text{)  -  co}}{{\text{t}}^{{\text{ - 1}}}}{\text{(  -  cot }}\dfrac{\pi}{{\text{6}}}{\text{)}} \hfill \\ \dfrac{\pi}{{\text{3}}}{\text{  -  co}}{{\text{t}}^{{\text{ - 1}}}}\left[ {{\text{cot }}\left( {{{\pi   -  }}\dfrac{\pi}{{\text{6}}}} \right)} \right] \hfill \\ \dfrac{\pi}{{\text{3}}}{\text{  -  co}}{{\text{t}}^{{\text{ - 1}}}}\left[ {{\text{cot }}\left( {\dfrac{{{{5\pi }}}}{{\text{6}}}} \right)} \right] \hfill \\ \dfrac{\pi}{{\text{3}}}{\text{  -  }}\dfrac{{{{5\pi }}}}{{\text{6}}}{\text{  =  }}\dfrac{{{{2\pi   -  5\pi }}}}{{\text{6}}}{\text{  =   -  }}\dfrac{{{{3\pi }}}}{{\text{6}}}{\text{  =   -  }}\dfrac{\pi}{{\text{2}}} \hfill \\  \end{align} $

(B) सही विकल्प है 

प्रश्नावली  A2

निम्नलिखित के मान ज्ञात कीजिए

1: $\mathbf{{\cos ^{{\text{ - 1}}}}\left( {{\text{cos }}\dfrac{{{{13\pi }}}}{{\text{6}}}} \right)}$ 

उत्तर: दिया गया है ${\cos ^{{\text{ - 1}}}}\left( {{\text{cos }}\dfrac{{{{13\pi }}}}{{\text{6}}}} \right)$

हमे ज्ञात है कि ${\text{x}} \in \left[ {{0, \pi }} \right]$ के लिए ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos x }}} \right){\text{  =  x}}$  

$\begin{align} {\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{13\pi }}}}{{\text{6}}}} \right){\text{  =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\;\left( {{{2\pi   +  }}\dfrac{\pi}{{\text{6}}}} \right)} \right) \hfill \\ {\text{                         =  co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos }}\left( {\dfrac{\pi}{{\text{6}}}} \right)} \right){\text{  =  }}\dfrac{\pi}{{\text{6}}} \hfill \\  \end{align} $

2: $\mathbf{{\tan ^{{\text{ - 1}}}}\left( {{\text{tan }}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)}$

उत्तर: दिया गया है ${\tan ^{{\text{ - 1}}}}\left( {{\text{tan }}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)$

हमे ज्ञात है कि ${\text{x}} \in \left[ {\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ के लिए ${\tan ^{{\text{ - 1}}}}\left( {\tan {\text{ x }}} \right){\text{  =  x}}$

$\begin{align} {\tan ^{{\text{ - 1}}}}\left( {\tan {\text{ }}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right){\text{  =  }}{\tan ^{{\text{ - 1}}}}\left( {\tan {\text{ }}\left( {{{\pi   +  }}\dfrac{\pi}{{\text{6}}}} \right)} \right) \hfill \\ {\text{                          =  }}{\tan ^{{\text{ - 1}}}}\left( {\tan {\text{ }}\left( {\dfrac{\pi}{{\text{6}}}} \right)} \right){\text{  =  }}\dfrac{\pi}{{\text{6}}} \hfill \\  \end{align} $

3: $\mathbf{{\text{2si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{24}}}}{{\text{7}}}}$

उत्तर: ${\text{2si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  =  2ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\sqrt {{{\text{5}}^{\text{2}}}{\text{  -  }}{{\text{3}}^{\text{2}}}} }}\;\;\;\;\;\;\;\;\;\;\quad \left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\text{h}}}{\text{ = }}} \right.\left. {\begin{array}{*{20}{c}} {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\sqrt {{{\text{h}}^{\text{2}}}{\text{ - }}{{\text{p}}^{\text{2}}}} }}}  \end{array}} \right]{\text{ }}$

${\text{ =  2ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}}$

$\begin{align} {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2  X  }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1  -  }}{{\left( {\dfrac{{\text{3}}}{{\text{4}}}} \right)}^{\text{2}}}}}\quad \;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1  -  }}{{\text{x}}^{\text{2}}}}}} \right] \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{{\text{3}}}{{\text{2}}}}}{{\dfrac{{\text{7}}}{{{\text{16}}}}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{24}}}}{{\text{7}}} \hfill \\  \end{align} $

4: $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{8}}}{{{\text{17}}}}{\text{  +  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{77}}}}{{{\text{36}}}}}$

उत्तर: ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{8}}}{{{\text{17}}}}{\text{  +  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{8}}}{{\sqrt {{\text{1}}{{\text{7}}^{\text{2}}}{\text{ - }}{{\text{8}}^{\text{2}}}} }}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\sqrt {{{\text{5}}^{\text{2}}}{\text{ - }}{{\text{3}}^{\text{2}}}} }}$

$\begin{align} \left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\text{h}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\sqrt {{{\text{h}}^{\text{2}}}{\text{  -  }}{{\text{p}}^{\text{2}}}} }}} \right] \hfill \\ {\text{      =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{8}}}{{{\text{15}}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}} \hfill \\  \end{align} $

${\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{8}}}{{{\text{15}}}}{\text{  +  }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1  -  }}\dfrac{{\text{8}}}{{{\text{15}}}}{\text{  X  }}\dfrac{{\text{3}}}{{\text{4}}}}}} \right)$

$\begin{align} \left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  +  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x  +  y}}}}{{{\text{1  -  x  X  y}}}}} \right)} \right] \hfill \\ {\text{                     =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{77}}}}{{{\text{36}}}} \hfill \\  \end{align} $

5: $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{{\text{5}}}{\text{  +  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{13}}}}{\text{  =  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{33}}}}{{{\text{65}}}}}$

उत्तर: ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{{\text{5}}}{\text{  +  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{13}}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{{\text{5}}^{\text{2}}}{\text{  -  }}{{\text{4}}^{\text{2}}}} }}{{\text{4}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1}}{{\text{3}}^{\text{2}}}{\text{  -  1}}{{\text{2}}^{\text{2}}}} }}{{{\text{12}}}}$

$\begin{align} {\text{            =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{12}}}} \hfill \\ {\text{           =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{5}}}{{{\text{12}}}}{\text{  +  }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1  -  }}\dfrac{{\text{5}}}{{{\text{12}}}}{\text{  X  }}\dfrac{{\text{3}}}{{\text{4}}}}}} \right) \hfill \\ \left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  +  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x  +  y}}}}{{{\text{1  -  x  X  y}}}}} \right)} \right] \hfill \\  \end{align} $

$\begin{align} {\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{56}}}}{{{\text{ 33 }}}} \hfill \\ {\text{ =  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{33}}}}{{\sqrt {{\text{5}}{{\text{6}}^{\text{2}}}{\text{  +  3}}{{\text{3}}^{\text{2}}}} }} \hfill \\ {\text{ =  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{33}}}}{{{\text{65}}}} \hfill \\  \end{align} $

6: $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{13}}}}{\text{  +  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  =  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{56}}}}{{{\text{65}}}}}$

उत्तर: ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{13}}}}{\text{  +  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1}}{{\text{3}}^{\text{2}}}{\text{  -  1}}{{\text{2}}^{\text{2}}}} }}{{{\text{12}}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\sqrt {{{\text{5}}^{\text{2}}}{\text{  -  }}{{\text{3}}^{\text{2}}}} }}$

$\begin{align} {\text{ =  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{56}}}}{{\sqrt {{\text{5}}{{\text{6}}^{\text{2}}}{\text{  +  3}}{{\text{3}}^{\text{2}}}} }} \hfill \\ {\text{ =  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{56}}}}{{{\text{65}}}} \hfill \\  \end{align} $ 

${\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{12}}}}$

$\begin{align} {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{5}}}{{{\text{12}}}}{\text{  +  }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1  -  }}\dfrac{{\text{5}}}{{{\text{12}}}}{\text{  X  }}\dfrac{{\text{3}}}{{\text{4}}}}}} \right) \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{56}}}}{{{\text{33}}}}} \right) \hfill \\  \end{align} $

7: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{63}}}}{{{\text{16}}}}{\text{  =  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{13}}}}{\text{  +  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}}$

उत्तर: ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{13}}}}{\text{  +  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{\sqrt {{\text{1}}{{\text{3}}^{\text{2}}}{\text{  -  }}{{\text{5}}^{\text{2}}}} }}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{{\text{5}}^{\text{2}}}{\text{  -  }}{{\text{3}}^{\text{2}}}} }}{{\text{3}}}{\text{ }}$

${\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{12}}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{{\text{3}}}$

$\begin{align} {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{5}}}{{{\text{12}}}}{\text{  +  }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1  -  }}\dfrac{{\text{5}}}{{{\text{12}}}}{\text{  X  }}\dfrac{{\text{3}}}{{\text{4}}}}}} \right) \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{63}}}}{{16}}} \right) \hfill \\ \end{align} $

8:$\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{8}}}{\text{  =  }}\dfrac{\pi}{{\text{4}}}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{8}}}$

$\begin{align} {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{1}}}{{\text{5}}}{\text{  +  }}\dfrac{{\text{1}}}{{\text{7}}}}}{{{\text{1  -  }}\dfrac{{\text{1}}}{{\text{5}}}{\text{  X  }}\dfrac{{\text{1}}}{{\text{7}}}}}} \right){\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{1}}}{{\text{8}}}{\text{  +  }}\dfrac{{\text{1}}}{{\text{3}}}}}{{{\text{1  - }}\dfrac{{\text{1}}}{{\text{8}}}{\text{  X  }}\dfrac{{\text{1}}}{{\text{3}}}}}} \right) \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{34}}}}{\text{  +  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{11}}}}{{{\text{23}}}} \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{{\text{12}}}}{{{\text{34}}}}{\text{  +  }}\dfrac{{{\text{11}}}}{{{\text{23}}}}}}{{{\text{1  -  }}\dfrac{{{\text{12}}}}{{{\text{34}}}}{\text{  X  }}\dfrac{{{\text{11}}}}{{{\text{23}}}}}}} \right) \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{650}}}}{{{\text{650}}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1)}} \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{\pi}{{\text{4}}}} \right){\text{  =  }}\dfrac{\pi}{{\text{4}}} \hfill \\  \end{align} $

सिद्ध कीजिए 

9: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1  -  x}}}}{{{\text{1  +  x}}}}} \right){\text{,   x}} \in {\text{[0, 1]}}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{  X  2ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} $

$\begin{align} {\text{ =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1  +  (}}\sqrt {\text{x}} {{\text{)}}^{\text{2}}}}}{{{\text{1  -  (}}\sqrt {\text{x}} {{\text{)}}^{\text{2}}}}}\;\;\;\;\;\;\;\;\;\quad \left[ {{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{  =  co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1  +  }}{{\text{x}}^{\text{2}}}}}{{{\text{1  -  }}{{\text{x}}^{\text{2}}}}}} \right] \hfill \\ {\text{ =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1  -  x}}}}{{{\text{1  +  x}}}}} \right) \hfill \\  \end{align} $ 

10: $\mathbf{{\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1  +  sin x}}} {\text{  +  }}\sqrt {{\text{1  -  sin x}}} }}{{\sqrt {{\text{1  +  sinx }}} {\text{ -  }}\sqrt {{\text{1  -  sinx}}} }}} \right){\text{  =  }}\dfrac{{\text{x}}}{{\text{2}}}{\text{,  x}} \in \left( {{\text{0,}}\dfrac{\pi}{{\text{4}}}} \right)}$

उत्तर: ${\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1  +  sin x}}} {\text{  +  }}\sqrt {{\text{1  -  sin x}}} }}{{\sqrt {{\text{1  +  sin x}}} {\text{  -  }}\sqrt {{\text{1  -  sin x}}} }}} \right)$

$\begin{align} {\text{ = co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  +  si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  +  2cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}} {\text{  +  }}\sqrt {{\text{co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  +  si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  -  2 cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}} }}{{\sqrt {{\text{co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  +  si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  +  2cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}} {\text{  -  }}\sqrt {{\text{co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  +  si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  -  2 cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}} }}} \right) \hfill \\ {\text{ =  co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{\left( {{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  +  sin}}\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{  +  }}\left( {{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  -  sin}}\dfrac{{\text{x}}}{{\text{2}}}} \right)}}{{\left( {{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  +  sin}}\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{  -  }}\left( {{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{  -  sin}}\dfrac{{\text{x}}}{{\text{2}}}} \right)}}} \right) \hfill \\ {\text{ =  co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2cos}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{2sin}}\dfrac{{\text{x}}}{{\text{2}}}}}} \right){\text{  =  co}}{{\text{t}}^{{\text{ - 1}}}}\left( {{\text{cot}}\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{  =  }}\dfrac{{\text{x}}}{{\text{2}}} \hfill \\  \end{align} $    

11: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1  +  x}}} {\text{  -  }}\sqrt {{\text{1  -  x}}} }}{{\sqrt {{\text{1  +  x}}} {\text{  +  }}\sqrt {{\text{1  -  x}}} }}} \right){\text{  =  }}\dfrac{\pi}{{\text{4}}}{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x ,  -  }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }} \leqslant {\text{x}} \leqslant {\text{1 }}}$

उत्तर:    दिया गया है ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1  +  x}}} {\text{  -  }}\sqrt {{\text{1  -  x}}} }}{{\sqrt {{\text{1  +  x}}} {\text{  +  }}\sqrt {{\text{1  -  x}}} }}} \right)$

मान लेते है कि ${{x  =  cos t }}\; \Rightarrow {\text{t  =  co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}$

${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1  +  x}}} {\text{ - }}\sqrt {{\text{1  -  x}}} }}{{\sqrt {{\text{1  +  x}}} {\text{  +  }}\sqrt {{\text{1  -  x}}} }}} \right){\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1  +  cos t}}} {\text{  -  }}\sqrt {{\text{1  -  cos t}}} }}{{\sqrt {{\text{1  +  cos t}}} {\text{  +  }}\sqrt {{\text{1  -  cos t}}} }}} \right)$

${\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{2 co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{t}}}{{\text{2}}}} {\text{  -  }}\sqrt {{\text{2 si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{t}}}{{\text{2}}}} }}{{\sqrt {{\text{2co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{t}}}{{\text{2}}}} {\text{  +  }}\sqrt {{\text{2si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{t}}}{{\text{2}}}} }}} \right){\text{  }}$

$\begin{align} {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{cos}}\dfrac{{\text{t}}}{{\text{2}}}{\text{  -  sin}}\dfrac{{\text{t}}}{{\text{2}}}}}{{\left. {{\text{cos}}\dfrac{{\text{t}}}{{\text{2}}}{\text{ + sin}}\dfrac{{\text{t}}}{{\text{2}}}} \right)}}{\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1  -  tan}}\dfrac{{\text{t}}}{{\text{2}}}}}{{{\text{1  +  tan}}\dfrac{{\text{t}}}{{\text{2}}}}}} \right)} \right. \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{tan}}\dfrac{\pi}{{\text{4}}}{\text{  -  tan}}\dfrac{{\text{t}}}{{\text{2}}}}}{{{\text{1  +  tan}}\dfrac{\pi}{{\text{4}}}{\text{tan}}\dfrac{{\text{t}}}{{\text{2}}}}}} \right){\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\left( {\dfrac{\pi}{{\text{4}}}{\text{  -  }}\dfrac{{\text{t}}}{{\text{2}}}} \right)} \right){\text{  =  }}\dfrac{\pi}{{\text{4}}}{\text{  -  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \end{align} $

12: $\mathbf{\dfrac{{{{9\pi }}}}{{\text{8}}}{\text{  -  }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}{\text{  =  }}\dfrac{{\text{9}}}{{\text{4}}}{\text{ si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2}}\sqrt {\text{2}} }}{{\text{3}}}}$

उत्तर:    $\dfrac{{{{9\pi }}}}{{\text{8}}}{\text{  -  }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}{\text{  =  }}\dfrac{{\text{9}}}{{\text{4}}}\left( {\dfrac{\pi}{{\text{2}}}{\text{  -  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}} \right)$

$\begin{align} {\text{ =  }}\dfrac{{\text{9}}}{{\text{4}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}\;\;\;\;\;\;\;\quad \left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  +  co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x  =  }}\dfrac{\pi}{{\text{2}}}} \right] \hfill \\ {\text{ =  }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{{\text{3}}^{\text{2}}}{\text{  -  }}{{\text{1}}^{\text{2}}}} }}{{\text{3}}} \hfill \\ {\text{ =  }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {\text{8}} }}{{\text{3}}}{\text{  =  }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2}}\sqrt {\text{2}} }}{{\text{3}}} \hfill \\  \end{align} $

निम्नलिखित को सरल कीजिए

13: $\mathbf{{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(cos x)  =  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(2cosec x)}}}$

उत्तर:  दिया गया है ${\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(cos x)  =  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(2cosec x)}}$

$\begin{align} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2cos x}}}}{{{\text{1  -  co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(2cosec x)}} \hfill \\ \left[ {{\text{2 ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1  -  }}{{\text{x}}^{\text{2}}}}}} \right] \hfill \\  \end{align} $

$\dfrac{{{\text{2cos x}}}}{{{\text{1  -  co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{\text{  =  2 cosec x}} \Rightarrow \dfrac{{{\text{2 cos x}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}}}{\text{  =  }}\dfrac{{\text{1}}}{{{\text{sin x}}}}$

$\begin{align} \Rightarrow \dfrac{{{\text{cos x}}}}{{{\text{sin x}}}}{\text{  =  1}} \Rightarrow {\text{cot x  =  cot}}\dfrac{\pi}{{\text{4}}} \hfill \\ \Rightarrow {\text{x  =  }}\dfrac{\pi}{{\text{4}}} \hfill \\  \end{align} $

14: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1  -  x}}}}{{{\text{1  +  x}}}}{\text{  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x   ,(x  >  0)}}}$

उत्तर:  दिया गया है ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1  -  x}}}}{{{\text{1  +  x}}}}{\text{  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

$\begin{align} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1  -  x}}}}{{{\text{1  +  x}}}}{\text{  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{1  -  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \Rightarrow \dfrac{\pi}{{\text{4}}}{\text{  =  }}\dfrac{{\text{3}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \Rightarrow \dfrac{\pi}{{\text{6}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ {\text{x  =  }}\dfrac{{\text{1}}}{{\sqrt {\text{3}} }} \hfill \\ \end{align} $

15: $\mathbf{{\text{sin }}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ ,| x |  <  1}}}$  बराबर होता है  

1. $\mathbf{\dfrac{{\text{x}}}{{\sqrt {{\text{1  -  }}{{\text{x}}^{\text{2}}}} }}}$

2. $\mathbf{\dfrac{{\text{1}}}{{\sqrt {{\text{1  -  }}{{\text{x}}^{\text{2}}}} }}}$

3. $\mathbf{\dfrac{{\text{1}}}{{\sqrt {{\text{1  +  }}{{\text{x}}^{\text{2}}}} }}}$

4. $\mathbf{\dfrac{{\text{x}}}{{\sqrt {{\text{1  +  }}{{\text{x}}^{\text{2}}}} }}}$

उत्तर: ${\text{sin}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ = }}$

$\begin{align} {\text{sin}}{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{\text{1  +  }}{{\text{x}}^{\text{2}}}} }}} \right)^{{\text{ - 1}}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\quad \left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\text{b}}}{\text{  =  si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\sqrt {{{\text{h}}^{\text{2}}}{\text{  +  b}}} }}} \right] \hfill \\ {\text{sin}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{  =  }}\dfrac{{\text{x}}}{{\sqrt {{\text{1  +  }}{{\text{x}}^{\text{2}}}} }} \hfill \\  \end{align} $

(D) सही उत्तर है 

16: $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1  -  x)  -  2 si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x  = }}\dfrac{\pi}{{\text{2}}}}$ का मान बराबर है

1. $\mathbf{{\text{0,}}\dfrac{{\text{1}}}{{\text{2}}}}$            

2. $\mathbf{{\text{1,}}\dfrac{{\text{1}}}{{\text{2}}}}$                

3. $\mathbf{{\text{0}}}$                    

4. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}}$

उत्तर: मान लेते है कि ${{x  =  sin t }} \Rightarrow {\text{ t  =  si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

${\text{(1  -  sin t)  =  sin}}\left( {\dfrac{\pi}{{\text{2}}}{\text{  +  2t}}} \right) \Rightarrow {\text{1  -  sin t  =  cos 2t}}$

 ${\text{1  -  sin t  =  1  -  2si}}{{\text{n}}^{\text{2}}}{\text{t}} \Rightarrow {\text{2}}{{\text{x}}^{\text{2}}}{\text{  -  x  =  0}} \Rightarrow {\text{x (2x  -  1)  =  0}}$

  ${{x  =  0, }}\dfrac{{\text{1}}}{{\text{2}}}$

(A) सही उत्तर है 

17: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\text{y}}}} \right){\text{  -  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x  -  y}}}}{{{\text{x  +  y}}}}}$  का मान  

1. $\mathbf{\dfrac{\pi}{{\text{2}}}}$                 

2. $\mathbf{\dfrac{\pi}{{\text{3}}}}$          

3. $\mathbf{\dfrac{\pi}{{\text{4}}}}$                   

4. $\mathbf{\dfrac{{{{3\pi }}}}{{\text{4}}}}$  

उत्तर:   (C) सही उत्तर है 

${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\text{y}}}} \right){\text{  -  ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x  -  y}}}}{{{\text{x  +  y}}}}{\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{x}}}{{\text{y}}}{\text{  -  }}\dfrac{{{\text{x  -  y}}}}{{{\text{x  +  y}}}}}}{{{\text{1  +  }}\dfrac{{\text{x}}}{{\text{y}}}{\text{  X  }}\dfrac{{{\text{x  -  y}}}}{{{\text{x  +  y}}}}}}} \right)$        

$\begin{align} {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x(x  +  y)  -  y(x  -  y)}}}}{{{\text{y(x  +  y)  +  x(x  -  y)}}}}} \right) \hfill \\ {\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\left. {{{\text{x}}^{\text{2}}}{\text{  +  xy  -  xy  +  }}{{\text{y}}^{\text{2}}}} \right)}}{{{{\text{x}}^{\text{2}}}{\text{  +  xy  -  xy  +  }}{{\text{y}}^{\text{2}}}}}} \right){\text{ =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\left. {{{\text{x}}^{\text{2}}}{\text{  +  }}{{\text{y}}^{\text{2}}}} \right)}}{{{{\text{x}}^{\text{2}}}{\text{  +  }}{{\text{y}}^{\text{2}}}}}} \right){\text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1)  =  }}\dfrac{\pi}{{\text{4}}} \hfill \\ \end{align} $

 (C) सही उत्तर है 

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions In Hindi

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