Inverse Trigonometric Functions Class 12 important questions with answers PDF download
Inverse trigonometry is one of the most important topics in the NCERT curriculum for Class 12 students. The free PDF solutions to Class 12 Maths Chapter 2 Important Questions on Inverse trigonometry are prepared by Vedantu experts according to the NCERT curriculum. These solutions are carefully prepared in such a way that it provides students with a step by step approach to solve any problems according to the Class 12 Maths Syllabus.
The Inverse Trigonometric Functions Class 12 Important Questions Covers Major Topics On Basic concepts and Properties of Inverse trigonometric functions.
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Access Class 12 Maths Chapter 2: Inverse Trigonometric Functions Important Questions
Very Short Questions and Answers (1 Marks Questions)
1. Write the Principal Value.
(i) Write the Principal Value of ${{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$.
Ans: Let ${{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=x$
Taking sine on both sides we get,
$\sin \left( x \right)=-\dfrac{\sqrt{3}}{2}$
But, the principal value of $\sin \left( \dfrac{\pi }{3} \right)$ is $\dfrac{\sqrt{3}}{2}$ and we know that and $\sin \left( -x \right)=-\sin \left( x \right)$.
$\Rightarrow \sin \left( -\dfrac{\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}$
$\therefore x=-\dfrac{\pi }{3}$
Hence, the principal value of ${{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$ is $-\dfrac{\pi }{3}$.
(ii) Write the Principal Value of ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$.
Ans: Let ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=x$
Taking cosine on both sides we get,
$\cos \left( x \right)=\dfrac{\sqrt{3}}{2}$
But, the principal value of $\cos \left( \dfrac{\pi }{6} \right)$ is $\dfrac{\sqrt{3}}{2}$.
$\therefore x=\dfrac{\pi }{6}$
Hence, the principal value of ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$ is $\dfrac{\pi }{6}$.
(iii) Write the Principal Value of ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$.
Ans: Let ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)=x$
Taking tan on both sides we get,
$\tan \left( x \right)=-\dfrac{1}{\sqrt{3}}$
But, the principal value of $\tan \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{\sqrt{3}}$ and we know that and $\tan \left( -x \right)=-\tan \left( x \right)$.
$\Rightarrow \tan \left( -\dfrac{\pi }{6} \right)=-\dfrac{1}{\sqrt{3}}$
$\therefore x=-\dfrac{\pi }{6}$
Hence, the principal value of ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$ is $-\dfrac{\pi }{6}$.
(iv) Write the Principal Value of $\text{cose}{{\text{c}}^{-1}}\left( -2 \right)$.
Ans: Let $\text{cose}{{\text{c}}^{-1}}\left( -2 \right)=x$
Taking cosec on both sides we get,
$\text{cosec}\left( x \right)=-2$
But, the principal value of $\sin \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{2}$ i.e., the principal value of $\text{cosec}\left( \dfrac{\pi }{6} \right)$ is $2$ and we know that and $\text{cosec}\left( -x \right)=-\text{cosec}\left( x \right)$.
$\Rightarrow \text{cosec}\left( -\dfrac{\pi }{6} \right)=-2$
$\therefore x=-\dfrac{\pi }{6}$
Hence, the principal value of $\text{cose}{{\text{c}}^{-1}}\left( -2 \right)$ is $-\dfrac{\pi }{6}$.
(v) Write the Principal Value of $\text{co}{{\text{t}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$.
Ans: Let $\text{co}{{\text{t}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)=x$
Taking cot on both sides we get,
$\text{cot}\left( x \right)=\dfrac{1}{\sqrt{3}}$
But, the principal value of \[\text{tan}\left( \dfrac{\pi }{3} \right)\] is \[\sqrt{3}\] i.e., the principal value of $\text{cot}\left( \dfrac{\pi }{3} \right)$ is \[\dfrac{1}{\sqrt{3}}\].
$\Rightarrow \text{cot}\left( \dfrac{\pi }{3} \right)=\dfrac{1}{\sqrt{3}}$
$\therefore x=\dfrac{\pi }{3}$
Hence, the principal value of $\text{co}{{\text{t}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$ is $\dfrac{\pi }{3}$.
(vi) Write the Principal Value of $\text{se}{{\text{c}}^{-1}}\left( -2 \right)$.
Ans: Let $\text{se}{{\text{c}}^{-1}}\left( -2 \right)=x$
Taking sec on both sides we get,
$\text{sec}\left( x \right)=-2$
But, the principal value of \[\text{cos}\left( \dfrac{2\pi }{3} \right)\] is \[-\dfrac{1}{2}\] i.e., the principal value of $\text{sec}\left( \dfrac{2\pi }{3} \right)$ is \[-2\].
$\Rightarrow \text{sec}\left( \dfrac{2\pi }{3} \right)=-2$
$\therefore x=\dfrac{2\pi }{3}$
Hence, the principal value of $\text{se}{{\text{c}}^{-1}}\left( -2 \right)$ is $\dfrac{2\pi }{3}$.
(vii) Write the Principal Value of $\text{si}{{\text{n}}^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+\text{co}{{\text{s}}^{-1}}\left( -\dfrac{1}{2} \right)+\text{ta}{{\text{n}}^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$.
Ans: Let $\text{si}{{\text{n}}^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=x$
Taking sine on both sides we get,
$\sin \left( x \right)=-\dfrac{\sqrt{3}}{2}$
But, the principal value of $\sin \left( \dfrac{\pi }{3} \right)$ is $\dfrac{\sqrt{3}}{2}$ and we know that and $\sin \left( -x \right)=-\sin \left( x \right)$
$\Rightarrow \sin \left( -\dfrac{\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}$
$\therefore x=-\dfrac{\pi }{3}$
Let $\text{co}{{\text{s}}^{-1}}\left( -\dfrac{1}{2} \right)=y$
Taking cosine on both sides we get,
$\cos \left( y \right)=-\dfrac{1}{2}$
But, the principal value of \[\text{cos}\left( \dfrac{2\pi }{3} \right)\] is \[-\dfrac{1}{2}\].
$\Rightarrow \text{cos}\left( \dfrac{2\pi }{3} \right)=-\dfrac{1}{2}$
$\therefore y=\dfrac{2\pi }{3}$
Let ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)=z$
Taking tan on both sides we get,
$\tan \left( z \right)=-\dfrac{1}{\sqrt{3}}$
But, the principal value of $\tan \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{\sqrt{3}}$ and we know that and $\tan \left( -x \right)=-\tan \left( x \right)$.
$\Rightarrow \tan \left( -\dfrac{\pi }{6} \right)=-\dfrac{1}{\sqrt{3}}$
$\therefore z=-\dfrac{\pi }{6}$
Hence, the principal value of $\text{si}{{\text{n}}^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+\text{co}{{\text{s}}^{-1}}\left( -\dfrac{1}{2} \right)+\text{ta}{{\text{n}}^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$ is \[x+y+z\].
\[\Rightarrow -\dfrac{\pi }{3}+\dfrac{2\pi }{3}-\dfrac{\pi }{6}=\dfrac{\pi }{6}\].
2. Write the Value of the Function.
(i) What is the Value of the Function \[\text{ta}{{\text{n}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)\].
Ans: Let $\text{ta}{{\text{n}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)=x$
Taking tan on both sides we get,
$\tan \left( x \right)=\dfrac{1}{\sqrt{3}}$
But, the principal value of $\tan \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{\sqrt{3}}$.
$\Rightarrow \tan \left( \dfrac{\pi }{6} \right)=\dfrac{1}{\sqrt{3}}$
$\therefore x=\dfrac{\pi }{6}$
Let ${{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)=y$
Taking sec on both sides we get,
$\sec \left( y \right)=\dfrac{2}{\sqrt{3}}$
But, the principal value of \[\text{cos}\left( \dfrac{\pi }{6} \right)\] is \[\dfrac{\sqrt{3}}{2}\]
$\Rightarrow \sec \left( \dfrac{\pi }{6} \right)=\dfrac{2}{\sqrt{3}}$
$\therefore y=\dfrac{\pi }{6}$
Hence, the principal value of \[\text{ta}{{\text{n}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)\] is $x+y$.
$\Rightarrow \dfrac{\pi }{6}-\dfrac{\pi }{6}=0$.
(ii) What is the Value of the Function \[{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)-{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\].
Ans: Let ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=x$
Taking sine on both sides we get,
$\sin \left( x \right)=-\dfrac{1}{2}$
But, the principal value of $\sin \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{2}$ and we know that $\sin \left( -x \right)=-\sin \left( x \right)$.
$\Rightarrow \sin \left( -\dfrac{\pi }{6} \right)=-\dfrac{1}{2}$
$\therefore x=-\dfrac{\pi }{6}$
Let ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=y$
Taking cos on both sides we get,
$\cos \left( y \right)=\dfrac{\sqrt{3}}{2}$
But, the principal value of \[\text{cos}\left( \dfrac{\pi }{6} \right)\] is \[\dfrac{\sqrt{3}}{2}\]
\[\Rightarrow \text{cos}\left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}\]
$\therefore y=\dfrac{\pi }{6}$
Hence, the principal value of \[{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)-{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\] is $x-y$.
$\Rightarrow -\dfrac{\pi }{6}-\dfrac{\pi }{6}=-\dfrac{\pi }{3}$.
(iii) What is the Value of the Function \[{{\tan }^{-1}}\left( 1 \right)-{{\cot }^{-1}}\left( -1 \right)\].
Ans: Let ${{\tan }^{-1}}\left( 1 \right)=x$
Taking tan on both sides we get, $\tan \left( x \right)=1$
But, the principal value of $\tan \left( \dfrac{\pi }{4} \right)$ is $1$.
$\Rightarrow \tan \left( \dfrac{\pi }{4} \right)=1$
$\therefore x=\dfrac{\pi }{4}$
Let ${{\cot }^{-1}}\left( -1 \right)=y$
Taking cot on both sides we get, \[\cot \left( y \right)=-1\]
But we know that the principal value of $\cot \left( \dfrac{3\pi }{4} \right)=-1$.
\[\Rightarrow \cot \left( \dfrac{3\pi }{4} \right)=-1\]
$\therefore y=\dfrac{3\pi }{4}$
Hence, the principal value of \[{{\tan }^{-1}}\left( 1 \right)-{{\cot }^{-1}}\left( -1 \right)\] is $x-y$
$\Rightarrow \dfrac{\pi }{4}-\dfrac{3\pi }{4}=-\dfrac{\pi }{2}$.
(iv) What is the Value of the Function \[\text{cose}{{\text{c}}^{-1}}\left( \sqrt{2} \right)+{{\sec }^{-1}}\left( \sqrt{2} \right)\].
Ans: We know that the principal value of $\sin \left( \dfrac{\pi }{4} \right),\cos \left( \dfrac{\pi }{4} \right)$ is $\dfrac{1}{\sqrt{2}}$.
Let $\text{cose}{{\text{c}}^{-1}}\left( \sqrt{2} \right)=x$ and $\text{se}{{\text{c}}^{-1}}\left( \sqrt{2} \right)=y$
Taking cosec on both sides of first equation and sec on both sides of second equation we get, $\text{cosec}\left( x \right)=\sqrt{2}$ and $\sec \left( y \right)=\sqrt{2}$.
But, the principal value of $\sin \left( \dfrac{\pi }{4} \right),\cos \left( \dfrac{\pi }{4} \right)$ is $\dfrac{1}{\sqrt{2}}$ and $\text{cosec}\left( \text{x} \right)\text{=}\dfrac{\text{1}}{\text{sinx}}\text{,sec}\left( \text{x} \right)\text{=}\dfrac{\text{1}}{\text{cosx}}$.
$\text{cosec}\left( \dfrac{\pi }{4} \right)=\sqrt{2}$ and $\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}$.
$\therefore x=y=\dfrac{\pi }{4}$
Hence, the principal value of \[\text{cose}{{\text{c}}^{-1}}\left( \sqrt{2} \right)+{{\sec }^{-1}}\left( \sqrt{2} \right)\] is $x+y$.
$\Rightarrow \dfrac{\pi }{4}+\dfrac{\pi }{4}=\dfrac{\pi }{2}$.
(v) What is the Value of the Function \[\text{ta}{{\text{n}}^{-1}}\left( 1 \right)+{{\cot }^{-1}}\left( 1 \right)+{{\sin }^{-1}}\left( 1 \right)\].
Ans: Let ${{\tan }^{-1}}\left( 1 \right)=x$
Taking tan on both sides we get, $\tan \left( x \right)=1$
But, the principal value of $\tan \left( \dfrac{\pi }{4} \right)$ is $1$.
$\Rightarrow \tan \left( \dfrac{\pi }{4} \right)=1$
$\therefore x=\dfrac{\pi }{4}$
Let ${{\cot }^{-1}}\left( 1 \right)=y$
Using the trigonometric identity ${{\cot }^{-1}}\left( x \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( x \right)$ we get
$y=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( 1 \right)$
But, ${{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$ (proved above)
$\therefore y=\dfrac{\pi }{4}$
Let ${{\sin }^{-1}}\left( 1 \right)=z$
Taking sine on both sides we get, $\sin \left( z \right)=1$
But, the principal value of $\sin \left( \dfrac{\pi }{2} \right)$ is $1$.
$\Rightarrow \sin \left( \dfrac{\pi }{2} \right)=1$
$\therefore z=\dfrac{\pi }{2}$
Hence, the principal value of \[\text{ta}{{\text{n}}^{-1}}\left( 1 \right)+{{\cot }^{-1}}\left( 1 \right)+{{\sin }^{-1}}\left( 1 \right)\] is $x+y+z$.
$\Rightarrow \dfrac{\pi }{4}+\dfrac{\pi }{4}+\dfrac{\pi }{2}=\pi $.
(vi) What is the Value of the Function \[{{\sin }^{-1}}\left( \sin \dfrac{4\pi }{5} \right)\].
Ans: Let \[{{\sin }^{-1}}\left( \sin \dfrac{4\pi }{5} \right)=x\]
Taking sine on both sides we get, \[\sin \dfrac{4\pi }{5}=\sin x\].
Writing $\sin \left( \dfrac{4\pi }{5} \right)$ as $\sin \left( \pi -\dfrac{\pi }{5} \right)=\sin \left( \dfrac{\pi }{5} \right)$.
$\therefore x=\dfrac{\pi }{5}$
Hence, the principal value of \[{{\sin }^{-1}}\left( \sin \dfrac{4\pi }{5} \right)\] is \[\dfrac{\pi }{5}\].
(vii) What is the Value of the Function \[{{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)\].
Ans: Let \[{{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)=x\]
Taking tan on both sides we get, \[\tan \dfrac{5\pi }{6}=\tan x\]
Writing $\tan \left( \dfrac{5\pi }{6} \right)$ as $\tan \left( \pi -\dfrac{\pi }{6} \right)=-\tan \left( \dfrac{\pi }{6} \right)$. Also, we know that $\tan \left( -x \right)=-\tan \left( x \right)$.
$\therefore x=-\dfrac{\pi }{6}$
Hence, the principal value of \[{{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)\] is \[-\dfrac{\pi }{6}\].
(viii) What is the Value of the Function \[\text{cose}{{\text{c}}^{-1}}\left( \text{cosec}\dfrac{3\pi }{4} \right)\].
Ans: Let \[\text{cose}{{\text{c}}^{-1}}\left( \text{cosec}\dfrac{3\pi }{4} \right)\]
Taking cosec on both sides we get, \[\text{cosec}\dfrac{3\pi }{4}=\cos \text{ec}\left( x \right)\]
Writing \[\text{cosec}\left( \dfrac{3\pi }{4} \right)\] as $\text{cosec}\left( \pi -\dfrac{\pi }{4} \right)=\text{cosec}\left( \dfrac{\pi }{4} \right)$.
$\therefore x=\dfrac{\pi }{4}$
Hence, the principal value of \[\text{cose}{{\text{c}}^{-1}}\left( \text{cosec}\dfrac{3\pi }{4} \right)\] is \[\dfrac{\pi }{4}\].
Long Questions and Answers (4 Marks Questions)
3. Show that \[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}} \right)=\dfrac{\pi }{4}+\dfrac{x}{2}\], $x\in \left[ 0,\pi \right]$.
Ans: Using the trigonometric identities
$\sqrt{1+\cos x}=2\cos \left( \dfrac{x}{2} \right)$ and $\sqrt{1-\cos x}=2\sin \left( \dfrac{x}{2} \right)$ we get,
\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}} \right)={{\tan }^{-1}}\left( \dfrac{2\cos \left( \dfrac{x}{2} \right)+2\sin \left( \dfrac{x}{2} \right)}{2\cos \left( \dfrac{x}{2} \right)-2\sin \left( \dfrac{x}{2} \right)} \right)\] …..(1)
Dividing the numerator and denominator of (1) by $2\cos \left( \dfrac{x}{2} \right)$ we get,
\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}} \right)={{\tan }^{-1}}\left( \dfrac{1+\tan \left( \dfrac{x}{2} \right)}{1-\tan \left( \dfrac{x}{2} \right)} \right)\] …..(2)
Now using the identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on equation (2) we get,
\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}} \right)={{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( \tan \left( \dfrac{x}{2} \right) \right)\]
\[\Rightarrow \dfrac{\pi }{4}+\dfrac{x}{2}\]
4. Show that \[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)-{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)=\dfrac{\pi }{4}\], $x\in \left( 0,\dfrac{\pi }{2} \right)$.
Ans: Using the trigonometric identities
\[\sqrt{1+\cos x}=2\cos \left( \dfrac{x}{2} \right)\] and \[\sqrt{1-\cos x}=2\sin \left( \dfrac{x}{2} \right)\] on \[{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)\] we get,
\[{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)={{\cot }^{-1}}\left( \dfrac{2\cos \left( \dfrac{x}{2} \right)}{2\sin \left( \dfrac{x}{2} \right)} \right)\]
\[\Rightarrow {{\cot }^{-1}}\left( \cot \left( \dfrac{x}{2} \right) \right)\]
But we know that ${{\cot }^{-1}}\left( \cot x \right)=x$, therefore,
\[{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)={{\cot }^{-1}}\left( \cot \left( \dfrac{x}{2} \right) \right)\]
\[\Rightarrow \dfrac{x}{2}\]
Let it be known as equation (1)
Now let us solve \[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)\] ….. (2)
Using the identities \[1={{\sin }^{2}}\left( \dfrac{x}{2} \right)+{{\cos }^{2}}\left( \dfrac{x}{2} \right)\] and \[\sin x=2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)\] in the denominator of (2) we get,
\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\cos x}{{{\sin }^{2}}\left( \dfrac{x}{2} \right)+{{\cos }^{2}}\left( \dfrac{x}{2} \right)-2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)} \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\cos x}{{{\left( \cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right) \right)}^{2}}} \right)\]
Let it be known as equation (3).
Now using the identity \[\cos x={{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)\] in the numerator of (3) we get,
\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\left( \cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right) \right)\cdot \left( \cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right)}{{{\left( \cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right) \right)}^{2}}} \right)\] …..(4)
Cancelling the \[\left( \cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right) \right)\] term from numerator and denominator of (4) we get
\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right)} \right)\] …..(5)
Simplifying it further by taking $\cos \left( \dfrac{x}{2} \right)$ common from both numerator and denominator of (5) we get,
\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left( \dfrac{1+\tan \left( \dfrac{x}{2} \right)}{1-\tan \left( \dfrac{x}{2} \right)} \right)\] …..(6)
Now using the identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on (6) we get,
\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( \tan \left( \dfrac{x}{2} \right) \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)=\dfrac{\pi }{4}+\dfrac{x}{2}\]
Let it be known as equation (7).
Using equations (1) and (7) we get
\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)-{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)=\dfrac{\pi }{4}+\dfrac{x}{2}-\dfrac{x}{2}\]
\[\therefore {{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)-{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)=\dfrac{\pi }{4}\]
5. Show that
\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\sin }^{-1}}\left( \dfrac{x}{a} \right)={{\cos }^{-1}}\left( \dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)\]
Ans: Using the trigonometric substitution $x=a\sin \theta $ in \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)\] we get,
\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{a\sin \theta }{\sqrt{{{a}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta }} \right)\] …..(1)
Taking out $a$ common from the denominator of (1) and cancelling out with numerator we get
\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }} \right)\] …..(2)
Using the trigonometric identity $\sqrt{1-{{\sin }^{2}}x}=\cos x$ in the denominator of (2) we get,
\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{\sin \theta }{\cos \theta } \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \tan (\theta ) \right)\]
Let it be known as equation (3).
Using the result ${{\tan }^{-1}}\left( \tan x \right)=x$ in (3) we get
\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)=\theta \] …..(4)
Let us now re-substitute
$\begin{align}& x=a\sin \theta \\ & \Rightarrow \theta ={{\sin }^{-1}}\left(\dfrac{x}{a} \right) \\ \end{align}$
In equation (4) we get,
\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\sin }^{-1}}\left( \dfrac{x}{a} \right)\] …..(5)
Now, from $x=a\sin \theta $ we get
$\sin \theta =\dfrac{x}{a}$. Hence, using the trigonometric identity
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ we get,
${{\left( \dfrac{x}{a} \right)}^{2}}+{{\cos }^{2}}\theta =1$
$\Rightarrow {{\cos }^{2}}\theta =1-\dfrac{{{x}^{2}}}{{{a}^{2}}}$
$\Rightarrow \cos \theta =\sqrt{\dfrac{{{a}^{2}}-{{x}^{2}}}{{{a}^{2}}}}$
Let this be known as equation (6).
Taking ${{\cos }^{-1}}$ on both sides of (6) we get,
${{\cos }^{-1}}\left( \cos \theta \right)={{\cos }^{-1}}\left( \dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)$
$\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)$
From equation (4) we get,
\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\cos }^{-1}}\left( \dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)\].
6. Show that \[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]={{\tan }^{-1}}\left( \dfrac{300}{161} \right)\]
Ans: Let us first change \[{{\cos }^{-1}}\dfrac{8}{17}\] into the form of tan inverse.
Let \[{{\cos }^{-1}}\dfrac{8}{17}=x\]
Taking cosine on both sides we get
\[\cos \left( {{\cos }^{-1}}\dfrac{8}{17} \right)=\cos x\]
\[\Rightarrow \cos x=\dfrac{8}{17}\]
Let this be known as equation (1).
Now using the trigonometric identity ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ we get,
${{\sin }^{2}}x=1-{{\left( \dfrac{8}{17} \right)}^{2}}$
$\Rightarrow \sin x=\dfrac{15}{17}$
Let this be known as equation (2).
From (1) and (2),
$\tan x=\dfrac{\sin x}{\cos x}$
$\Rightarrow \tan x=\dfrac{15}{8}$
Let this be known as equation (3).
Taking ${{\tan }^{-1}}$ on both sides of equation (3) we get,
${{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}\left( \dfrac{15}{8} \right)$
$\Rightarrow x={{\tan }^{-1}}\left( \dfrac{15}{8} \right)$
Let this be known as equation (4).
Similarly let us now change \[{{\sin }^{-1}}\dfrac{8}{17}\] into the form of tan inverse.
Let \[{{\sin }^{-1}}\dfrac{8}{17}=y\]
Taking sine on both sides we get
\[\sin \left( {{\sin }^{-1}}\dfrac{8}{17} \right)=\sin y\]
\[\Rightarrow \sin y=\dfrac{8}{17}\]
Let this be known as equation (5).
Now using the trigonometric identity ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ on (5) we get,
${{\cos }^{2}}y=1-{{\left( \dfrac{8}{17} \right)}^{2}}$
$\Rightarrow \cos y=\dfrac{15}{17}$
Let this be known as equation (6).
From (5) and (6),
$\tan y=\dfrac{\sin y}{\cos y}$
$\Rightarrow \tan y=\dfrac{8}{15}$
Let this be known as equation (7).
Taking ${{\tan }^{-1}}$ on both sides of equation (7) we get,
${{\tan }^{-1}}\left( \tan y \right)={{\tan }^{-1}}\left( \dfrac{8}{15} \right)$
$\Rightarrow y={{\tan }^{-1}}\left( \dfrac{8}{15} \right)$
Let this be known as equation (8).
Substituting these values from (4) and (8) in the original equation we get,
\[\begin{align} & {{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right] \\ & \text{ }={{\cot }^{-1}}\left[ 2\tan \left( {{\tan }^{-1}}\dfrac{15}{8} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\tan }^{-1}}\dfrac{8}{15} \right) \right] \\ \end{align}\]
\[\Rightarrow {{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]={{\cot }^{-1}}\left( \dfrac{15}{4} \right)+{{\tan }^{-1}}\left(\dfrac{16}{15} \right)\]
Let this be known as equation (9).
Now using the trigonometric identity ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ on equation (9) we get,
\[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]={{\cot }^{-1}}\left( \dfrac{15}{4} \right)+\dfrac{\pi }{2}-{{\cot }^{-1}}\left( \dfrac{16}{15} \right)\]
Let this be known as equation (10).
Again, using the trigonometric identity ${{\cot }^{-1}}\left( \dfrac{xy+1}{y-x} \right)={{\cot }^{-1}}\left( x \right)-{{\cot }^{-1}}\left( y \right)$ on (10) we get,
\[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]=\dfrac{\pi }{2}+{{\cot }^{-1}}\left( \dfrac{5}{\dfrac{16}{15}-\dfrac{15}{4}} \right)\]
\[\Rightarrow {{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]=\dfrac{\pi }{2}+{{\cot }^{-1}}\left( \dfrac{300}{-161} \right)\]
Let this be known as equation (11).
Using ${{\cot }^{-1}}\left( x \right)=-{{\cot }^{-1}}\left( x \right)$ on (11) we get,
\[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]=\dfrac{\pi }{2}-{{\cot }^{-1}}\left( \dfrac{300}{161} \right)\] …..(12)
Again, using the trigonometric identity ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ on (12) we get,
\[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]={{\tan }^{-1}}\left( \dfrac{300}{161} \right)\]
Hence proved.
7. Show that \[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)=\dfrac{\pi }{4}+\dfrac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]
Ans: To solve this question, substitute ${{x}^{2}}=\cos \theta $, therefore,
\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }}{\sqrt{1+\cos \theta }-\sqrt{1-\cos \theta }} \right)\] …..(1)
Using the trigonometric identities $\sqrt{1+\cos x}=2\cos \left( \dfrac{x}{2} \right)$ and $\sqrt{1-\cos x}=2\sin \left( \dfrac{x}{2} \right)$ on equation (1) we get,
\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{2\cos \left( \dfrac{\theta }{2} \right)+2\sin \left( \dfrac{\theta }{2} \right)}{2\cos \left( \dfrac{\theta }{2} \right)-2\sin \left( \dfrac{\theta }{2} \right)} \right)\] …..(2)
Taking \[2\cos \left( \dfrac{\theta }{2} \right)\] common from numerator and denominator of (2) we get,
\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{1+\tan \left( \dfrac{\theta }{2} \right)}{1-\tan \left( \dfrac{\theta }{2} \right)} \right)\] …..(3)
Now using the identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on (3) we get,
\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( \tan \left( \dfrac{\theta }{2} \right) \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)=\dfrac{\pi }{4}+\dfrac{\theta }{2}\]
Let this be known as equation (4).
Re-substituting ${{x}^{2}}=\cos \theta $ in (4) we get,
\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)=\dfrac{\pi }{4}+\dfrac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]
Hence Proved.
8. Solve the following for $x$: \[{{\cot }^{-1}}2x+{{\cot }^{-1}}3x=\dfrac{\pi }{4}\]
Ans: To solve this question, use the identity ${{\cot }^{-1}}\left( \dfrac{xy-1}{x+y} \right)={{\cot }^{-1}}\left( x \right)+{{\cot }^{-1}}\left( y \right)$
\[{{\cot }^{-1}}2x+{{\cot }^{-1}}3x={{\cot }^{-1}}\left( \dfrac{6{{x}^{2}}-1}{5x} \right)\]
Hence,
\[{{\cot }^{-1}}\left( \dfrac{6{{x}^{2}}-1}{5x} \right)=\dfrac{\pi }{4}\] …..(1)
Taking cot on both sides we get,
\[\begin{align}& \cot \left( {{\cot }^{-1}}\left( \dfrac{6{{x}^{2}}-1}{5x} \right) \right)=\cot \left( \dfrac{\pi }{4} \right) \\ & \Rightarrow \dfrac{6{{x}^{2}}-1}{5x}=\cot \left( \dfrac{\pi }{4} \right) \\ \end{align}\]
Let it be known as equation (2).
Solving the equation (2) by substituting the principal value of $\cot \left( \dfrac{\pi }{4} \right)$ we get,
\[\begin{align}& 6{{x}^{2}}-1=5x \\ & \Rightarrow 6{{x}^{2}}-5x-1=0 \\ \end{align}\]
Simplifying it further we get,
\[\begin{align} & 6{{x}^{2}}+\left( -6x+x \right)-1=0 \\ & \left( 6x+1 \right)\left( x-1 \right)=0 \\ & \Rightarrow x=1,-\dfrac{1}{6} \\ \end{align}\]
But $x=-\dfrac{1}{6}$ is not possible. Therefore,
$x=1$.
9. Show that \[{{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)=\dfrac{\pi }{4},\text{ }m,n>0\]
Ans: To solve this problem use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}\left( x \right)-{{\tan }^{-1}}\left( y \right)$
Hence,
\[{{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{m}{n}-\dfrac{m-n}{m+n}}{1+\left( \dfrac{m-n}{m+n} \right)\left( \dfrac{m}{n} \right)} \right)\] …..(1)
Simplifying the fractions of equation (1) we get
\[{{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{{{m}^{2}}+{{n}^{2}}}{n\left( m+n \right)}}{1+\left( \dfrac{{{m}^{2}}-mn}{n\left( m+n \right)} \right)} \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)={{\tan }^{-1}}\left( \dfrac{{{m}^{2}}+{{n}^{2}}}{{{m}^{2}}+{{n}^{2}}} \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)={{\tan }^{-1}}\left( 1 \right)\]
Let this be known as equation (2).
But we know that,
$\tan \left( \dfrac{\pi }{4} \right)=1$
$\Rightarrow {{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$
Hence from equation (2) it is proved that \[{{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)=\dfrac{\pi }{4},\text{ }m,n>0\]
10. Prove that \[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\dfrac{x+y}{1-xy}\]
Ans: To solve this problem use the substitution,
x = tanθ
y = tanα
In the LHS of the given expression.
\[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha } \right) \right]\]
Let this be known as equation (1).
Using the trigonometric identity
\[\begin{align} & \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta \\ & \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\cos 2\theta \\ \end{align}\]
From (1) we get,
\[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \sin 2\theta \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \cos 2\alpha \right) \right]\]
\[\Rightarrow \tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\tan \left[ \theta +\alpha \right]\]
Let this be known as equation (2).
Now using the trigonometric identity, \[\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\] on equation (2) we get,
\[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\dfrac{\tan \theta +\tan \alpha }{1-\tan \theta \tan \alpha }\] …..(3)
Re-substituting $\theta ={{\tan }^{-1}}\left( x \right)$ and $\alpha ={{\tan }^{-1}}\left( y \right)$ in equation (3) we get,
\[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\dfrac{\tan \left( {{\tan }^{-1}}x \right)+\tan \left( {{\tan }^{-1}}y \right)}{1-\tan \left( {{\tan }^{-1}}x \right)\cdot \tan \left( {{\tan }^{-1}}y \right)}\]
\[\Rightarrow \tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\dfrac{x+y}{1-xy}\]
11. Solve the following for $x$: \[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\dfrac{2\pi }{3}\]
Ans: To solve this question, the substitution $x=\tan \theta $ in the LHS of the given equation
\[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)={{\cos }^{-1}}\left( \dfrac{{{\tan }^{2}}\theta -1}{{{\tan }^{2}}\theta +1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2\tan \theta }{1-{{\tan }^{2}}\theta } \right)\]
Let this be known as equation (1).
Using the trigonometric identities \[\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }=\tan 2\theta \] and \[\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\cos 2\theta \]
From (1) we get,
\[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)={{\cos }^{-1}}\left( -\cos 2\theta \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( -\tan 2\theta \right)\] …..(2)
But,
${{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}\left( x \right)$
${{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}\left( x \right)$
Therefore from (2) we get,
\[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\pi -{{\cos }^{-1}}\left( \cos 2\theta \right)-\dfrac{1}{2}{{\tan }^{-1}}\left( \tan 2\theta \right)\]
\[\Rightarrow {{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\pi -2\theta -\theta \]
\[\Rightarrow {{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\pi -3\theta \]
Let this be known as equation (3).
Re substituting $\theta ={{\tan }^{-1}}\left( x \right)$ in equation (3) we get,
\[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\pi -3{{\tan }^{-1}}x\] …..(4)
But it is given that \[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\dfrac{2\pi }{3}\]. Therefore from (4) we get,
\[\pi -3{{\tan }^{-1}}x=\dfrac{2\pi }{3}\]
\[\Rightarrow 3{{\tan }^{-1}}x=\dfrac{\pi }{3}\]
\[\Rightarrow x=\tan \left( \dfrac{\pi }{9} \right)\]
12. Prove that \[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}\]
Ans: To solve this problem use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ in the LHS of the given expression
Hence,
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{3}+\dfrac{1}{5}}{1-\dfrac{1}{15}} \right)+{{\tan }^{-1}}\left( \dfrac{\dfrac{1}{7}+\dfrac{1}{8}}{1-\dfrac{1}{56}} \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{4}{7} \right)+{{\tan }^{-1}}\left( \dfrac{3}{11} \right)\]
Let this be known as equation (1).
Again, using the trigonometric identity ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on (1) we get,
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{4}{7}+\dfrac{3}{11}}{1-\dfrac{12}{77}} \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{65}{65} \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( 1 \right)\]
Let this be known as equation (2).
But we know that,
$\tan \left( \dfrac{\pi }{4} \right)=1$
$\Rightarrow {{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$
Hence from equation (2) it is proved that \[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}\]
13. Solve the following for $x$: \[\tan \left( {{\cos }^{-1}}x \right)=\sin \left( {{\tan }^{-1}}2 \right),\text{ }x>0\]
Ans: To solve this question, write ${{\cos }^{-1}}x$ in LHS of the given expression in terms of tan inverse. Substitute,
\[{{\cos }^{-1}}x=\theta \]
\[\Rightarrow \cos \theta =x\]
Let this be known as equation (1).
Now using the trigonometric identity $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ we get from (1),
$\sin \theta =\sqrt{1-{{x}^{2}}}$ …..(2)
Hence from (1) and (2) we get,
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
$\Rightarrow \tan \theta =\dfrac{\sqrt{1-{{x}^{2}}}}{x}$
Let this be known as equation (3).
Taking tan inverse on both sides of equation (3) we get
${{\tan }^{-1}}\left( \tan \theta \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$
Let this be known as equation (4).
Hence from (4), we get the given equation as
\[\tan \left( {{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right) \right)=\sin \left( {{\tan }^{-1}}2 \right)\]
\[\Rightarrow \dfrac{\sqrt{1-{{x}^{2}}}}{x}=\sin \left( {{\tan }^{-1}}2 \right)\]
Let this be known as equation (5).
Now again write ${{\tan }^{-1}}2$ from RHS of equation (5) in terms of sine inverse. Substitute,
\[{{\tan }^{-1}}2=y\]
\[\Rightarrow \tan y=2\]
Let this be known as equation (6).
Now using the trigonometric identity $\sqrt{1+{{\tan }^{2}}\theta }=\sec \theta $ from (6) we get,
$\sec y=\sqrt{1+{{\left( 2 \right)}^{2}}}$
$\Rightarrow \sec y=\sqrt{5}$
$\Rightarrow \cos y=\dfrac{1}{\sqrt{5}}$
Let this be known as equation (7).
Using the trigonometric identity $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ we get from (7),
$\sin y=\sqrt{1-{{\left( \dfrac{1}{\sqrt{5}} \right)}^{2}}}$
$\Rightarrow \sin y=\dfrac{2}{\sqrt{5}}$
$\Rightarrow y={{\sin }^{-1}}\left( \dfrac{2}{\sqrt{5}} \right)$
Let this be known as equation (8).
Hence from equations (5), (6) and (8) we get,
\[\dfrac{\sqrt{1-{{x}^{2}}}}{x}=\dfrac{2}{\sqrt{5}}\] …..(9)
Squaring both sides and cross multiplying we get,
\[5\left( 1-{{x}^{2}} \right)=4{{x}^{2}}\]
\[\Rightarrow 5=9{{x}^{2}}\]
\[\Rightarrow x=\dfrac{\sqrt{5}}{3}\]
(Since $x>0$)
14. Prove that \[2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{32}{43} \right)\]
Ans: To solve this question, use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=2{{\tan }^{-1}}\left( x \right)$ in the LHS of the given expression.
\[2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{2\left( \dfrac{1}{5} \right)}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)\]
\[\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)\]
Let this be known as equation (1).
Now use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on equation (1) we get,
\[2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{5}{12}+\dfrac{1}{4}}{1-\dfrac{5}{48}} \right)\]
\[\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{32}{43} \right)\]
15. Evaluate \[\tan \left[ \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right) \right]\]
Ans: To solve this question, use the substitution \[\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right)=x\] in the LHS of the given expression. Therefore,
\[\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right)=x\]
\[\Rightarrow \cos 2x=\dfrac{3}{\sqrt{11}}\]
Let this be known as equation (1).
But we know that \[\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\cos 2x\]. Hence from (1) we get,
\[\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\dfrac{3}{\sqrt{11}}\] …..(2)
Applying the rule of component and divided on equation (2) we get,
\[\dfrac{1-{{\tan }^{2}}x+1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x-1-{{\tan }^{2}}x}=\dfrac{3+\sqrt{11}}{3-\sqrt{11}}\]
\[\Rightarrow \dfrac{2}{-2{{\tan }^{2}}x}=\dfrac{3+\sqrt{11}}{3-\sqrt{11}}\]
Let this be known as equation (3).
Taking reciprocal of equation (3) we get,
\[-{{\tan }^{2}}x=\dfrac{3-\sqrt{11}}{3+\sqrt{11}}\]
\[\Rightarrow {{\tan }^{2}}x=\dfrac{\sqrt{11}-3}{3+\sqrt{11}}\]
\[\Rightarrow \tan x=\sqrt{\dfrac{\sqrt{11}-3}{3+\sqrt{11}}}\]
Let this be known as equation (4).
Now re-substituting \[\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right)=x\] in (4) we get,
\[\tan \left[ \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right) \right]=\sqrt{\dfrac{\sqrt{11}-3}{3+\sqrt{11}}}\]
16. Prove that \[{{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)={{\tan }^{-1}}\left( \dfrac{a}{b} \right)-x\]
Ans: To solve this question, take $b\cos x$ common from both numerator and denominator from the LHS of the given equation.
\[{{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{a}{b}-\tan x}{1+\dfrac{a}{b}\tan x} \right)\] …..(1)
Now use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}\left( x \right)-{{\tan }^{-1}}\left( y \right)$ on equation (1) we get,
\[{{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)={{\tan }^{-1}}\left( \dfrac{a}{b} \right)-{{\tan }^{-1}}\left( \tan x \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)={{\tan }^{-1}}\left( \dfrac{a}{b} \right)-x\]
17. Prove that \[\cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}+{{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)=\pi \], $x>0$
Ans: To solve this question, use the trigonometric identity \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\] on \[\cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}\] . Therefore,
\[\cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}=\cot \left\{ \dfrac{\pi }{2}-{{\cot }^{-1}}x+\dfrac{\pi }{2}-{{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right\}\]
\[\Rightarrow \cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}=\cot \left\{ \pi -\left[ {{\cot }^{-1}}x+{{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right] \right\}\]
Let this be known as equation (1).
Now using the trigonometric identity \[{{\cot }^{-1}}\left( \dfrac{xy-1}{x+y} \right)={{\cot }^{-1}}\left( x \right)+{{\cot }^{-1}}\left( y \right)\] on equation (1) we get,
\[\cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}=\cot \left\{ \pi -{{\cot }^{-1}}\left( \dfrac{x\cdot \dfrac{1}{x}-1}{x+\dfrac{1}{x}} \right) \right\}\]
\[\Rightarrow \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}=\cot \left\{ \pi -\left[ {{\cot }^{-1}}\left( 0 \right) \right] \right\}\]
Let this be known as equation (2).
But we know that the principal value of
$\cot \left( \dfrac{\pi }{2} \right)=0$
$\Rightarrow {{\cot }^{-1}}\left( 0 \right)=\dfrac{\pi }{2}$
Hence, from equation (2) we get,
\[\cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}=\cot \left( \dfrac{\pi }{2} \right)\]
\[\Rightarrow \cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}=0\]
Let this be known as equation (3).
Also, we know that ${{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}\left( x \right)$. Therefore,
\[{{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)={{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+{{\cos }^{-1}}\left( -\left( -2{{x}^{2}}+1 \right) \right)\]
\[\Rightarrow {{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)={{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+\pi -{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)\]
\[\Rightarrow {{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)=\pi \]
Let this be known as equation (4).
Hence from equations (3) and (4) we get,
\[\cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}+{{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)=\pi \]
18. Prove that \[{{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)+{{\tan }^{-1}}\left( \dfrac{b-c}{1+bc} \right)+{{\tan }^{-1}}\left( \dfrac{c-a}{1+ac} \right)=0\] where $a,b,c>0$
Ans: To solve this question, use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}\left( x \right)-{{\tan }^{-1}}\left( y \right)$ on the LHS of the given expression. Therefore,
\[{{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)+{{\tan }^{-1}}\left( \dfrac{b-c}{1+bc} \right)+{{\tan }^{-1}}\left( \dfrac{c-a}{1+ac} \right)\]
\[\text{ }=\left( {{\tan }^{-1}}\left( a \right)-{{\tan }^{-1}}\left( b \right) \right)+\left( {{\tan }^{-1}}\left( b \right)-{{\tan }^{-1}}\left( c \right) \right)+\left( {{\tan }^{-1}}\left( c \right)-{{\tan }^{-1}}\left( a \right) \right)\]
\[\text{ }=0\]
19. Solve the following for $x$: \[2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\text{cosec }x \right)\]
Ans: To solve this question, use the trigonometric identity, $2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ on the LHS of the given expression.
\[2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( \dfrac{2\cos x}{1-{{\cos }^{2}}x} \right)\] …..(1)
Using the trigonometric identity ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ in the denominator of (1) we get,
\[2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( \dfrac{2\cos x}{{{\sin }^{2}}x} \right)\] …..(2)
Substituting the value from equation (2) into the given expression we get,
\[{{\tan }^{-1}}\left( \dfrac{2\cos x}{{{\sin }^{2}}x} \right)={{\tan }^{-1}}\left( 2\text{cosec }x \right)\] …..(3)
Taking tan on both the sides of equation (3) we get,
\[\tan \left( {{\tan }^{-1}}\left( \dfrac{2\cos x}{{{\sin }^{2}}x} \right) \right)=\tan \left( {{\tan }^{-1}}\left( 2\text{cosec }x \right) \right)\]
\[\Rightarrow \dfrac{2\cos x}{{{\sin }^{2}}x}=2\text{cosec }x\]
\[\Rightarrow \dfrac{\cos x}{{{\sin }^{2}}x}=\dfrac{1}{\sin x}\]
\[\Rightarrow \cos x=\sin x\]
It is possible only when $x=\dfrac{\pi }{4}$.
20. Express \[{{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)\] in the simplest form.
Ans: To solve this question, use the substitution $x=\sin \theta $ and $\sqrt{x}=\sin \phi $ . Therefore,
\[{{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)={{\sin }^{-1}}\left( \sin \theta \sqrt{1-{{\sin }^{2}}\phi }-\sin \phi \sqrt{1-{{\sin }^{2}}\theta } \right)\]
\[\Rightarrow {{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)={{\sin }^{-1}}\left( \sin \theta \cos \phi -\sin \phi \cos \theta \right)\]
Let this be known as equation (1).
But we know that $\sin a-\sin b=\sin a\cos b-\cos a\sin b$. Hence from (1) we get,
\[{{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)={{\sin }^{-1}}\left( \sin \left( \theta -\phi \right) \right)\]
\[\Rightarrow {{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)=\theta -\phi \]
Let this be known as equation (2).
Re-substituting $x=\sin \theta $ and $\sqrt{x}=\sin \phi $ in equation (2) we get,
\[{{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)={{\sin }^{-1}}\left( x \right)-{{\sin }^{-1}}\left( \sqrt{x} \right)\]
21. If \[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)=\pi \] then prove that $a+b+c=abc$.
Ans: To solve this question, use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on the LHS of the given expression. Therefore,
\[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)\] …..(1)
Again, using the same trigonometric identity on the LHS of the given expression and using (1), we get,
\[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)+{{\tan }^{-1}}\left( c \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{a+b}{1-ab}+c}{1-\dfrac{c\left( a+b \right)}{1-ab}} \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)={{\tan }^{-1}}\left( \dfrac{a+b+c-abc}{1-ab-ca-bc} \right)\]
Let this be known as equation (2).
But it is given that \[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)=\pi \]. Therefore, from equation (2) we get,
\[{{\tan }^{-1}}\left( \dfrac{a+b+c-abc}{1-ab-ca-bc} \right)=\pi \] …..(3)
Taking tan on both sides of the equation (3) we get,
\[\tan \left( {{\tan }^{-1}}\left( \dfrac{a+b+c-abc}{1-ab-ca-bc} \right) \right)=\tan \left( \pi \right)\]
\[\Rightarrow \dfrac{a+b+c-abc}{1-ab-ca-bc}=0\]
\[\Rightarrow a+b+c-abc=0\]
Hence proved that if \[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)=\pi \] then $a+b+c=abc$.
22. If \[{{\sin }^{-1}}x>{{\cos }^{-1}}x\] then \[x\] belong to which category?
Ans: To solve this question, use the trigonometric identity, ${{\sin }^{-1}}\left( x \right)+{{\cos }^{-1}}\left( x \right)=\dfrac{\pi }{2}$.
\[{{\sin }^{-1}}x>{{\cos }^{-1}}x\]
\[\Rightarrow {{\sin }^{-1}}x>\dfrac{\pi }{2}-{{\sin }^{-1}}x\]
\[\Rightarrow 2{{\sin }^{-1}}x>\dfrac{\pi }{2}\]
\[\Rightarrow {{\sin }^{-1}}x>\dfrac{\pi }{4}\]
Let this be known as expression (1).
Taking sine on both the sides of expression (1) we get,
\[\sin \left( {{\sin }^{-1}}x \right)>\sin \left( \dfrac{\pi }{4} \right)\]
\[\Rightarrow x>\sin \left( \dfrac{\pi }{4} \right)\]
\[\Rightarrow x>\dfrac{1}{\sqrt{2}}\]
Let this be known as expression (2).
Also, we know that $-1\le x\le 1$ (since range of $\sin $ and $\cos $ is $\left[ -1,1 \right]$. …..(3)
Hence from (2) and (3), the interval in which $x$ belongs is $\left( \dfrac{1}{\sqrt{2}},1 \right]$.
Inverse Trigonometric Functions Class 12 Important Questions - Free PDF Download
Chapter 2 - Inverse Trigonometry
Introduction
The opposite operations that the sine, cosine, tangent, secant, cosecant, and cotangent perform are provided by the inverse trigonometric functions. They are used in a right triangle to find the measure of an angle when two of the three side lengths are identified.
Principal Inverse Trigonometric Functions with Domain and Range
Functions | Domain | Range (Principal Value Branches) |
$y = \sin^{-1}x$ | [-1, 1] | $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ |
$y = \cos^{-1}x$ | [-1, 1] | $\left[0, \pi\right]$ |
$y = \text{cosec}^{-1}x$ | $\mathbb{R}$ - (-1, 1) | $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - {0}$ |
$y = \sec^{-1}x$ | $\mathbb{R}$ - (-1, 1) | $\left[0, \pi\right] - \dfrac{\pi}{2}$ |
$y = \tan^{-1}x$ | $\mathbb{R}$ | $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ |
$y = \cot^{-1}x$ | $\mathbb{R}$ | $\left(0, \pi\right)$ |
Properties of Inverse Trigonometric Functions
In order to not only solve problems but also to provide a deeper understanding of this idea, the properties of the 6 inverse trigonometric functions are important. The properties of inverse trigonometric relations are nothing but the relationship between the 6 fundamental trigonometric functions.
Here is the representation of important properties involving inverse trigonometric functions for quick reference:
Property | Formula/Condition |
Relation between inverse trigonometric functions: |
|
Odd Function Property : |
|
Sum Identities: |
|
Addition of tan⁻¹ Values | $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\dfrac{x + y}{1 - xy}\right), \; xy < 1$ |
Subtraction of tan⁻¹ Values | $\tan^{-1}(y) = \tan^{-1}\left(\dfrac{x - y}{1 + xy}\right), \; xy > -1$ |
Double Angle (sin⁻¹) | $2\tan^{-1}(x) = \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)$ ; |
Double Angle (cos⁻¹) | $2\tan^{-1}(x) = \cos^{-1}\left(\dfrac{1 - x^2}{1 + x^2}\right), \; x \geq 0$ |
Double Angle (tan⁻¹) | $2\tan^{-1}(x) = \tan^{-1}\left(\dfrac{2x}{1 - x^2}\right), \; -1 < x < 1$ |
The solution to Class 12 Maths Chapter 2 Important Questions provided by Vedantu has steps to derive the properties of the above-mentioned inverse trigonometric functions.
Benefits of Class 12 Science Chapter 2 Inverse Trigonometric Functions
Foundation for Advanced Mathematics: Inverse trigonometric functions are a cornerstone for higher-level math concepts like calculus, integration, and differential equations, which are critical in engineering, physics, and other technical fields.
Real-Life Applications: Understanding this chapter helps in solving real-world problems involving angles, distances, and periodic phenomena, such as navigation, signal processing, and wave mechanics.
Improved Problem-Solving Skills: By mastering the properties and formulas, students can tackle complex trigonometric and algebraic problems more effectively, enhancing their analytical abilities.
Competitive Exam Preparation: Topics in this chapter frequently appear in competitive exams like JEE, NEET, and other entrance tests, making it essential for scoring well.
Conceptual Clarity: Learning about the relationships between different trigonometric functions and their inverses sharpens conceptual understanding, aiding in better comprehension of subsequent chapters.
Tips to Study Class 12 Maths Chapter 2 Inverse Trigonometric Functions
Understand Basic Trigonometric Functions: Before diving into inverse trigonometric functions, ensure a strong grasp of basic trigonometric concepts and their properties, as they form the foundation for this chapter.
Memorise Key Formulas: Create a formula sheet for inverse trigonometric properties, such as addition/subtraction formulas and relationships between functions, and revise it regularly for quick recall.
Learn Domain and Range: Pay close attention to the domain and range of each inverse trigonometric function. This understanding is critical when solving equations and graphing functions.
Practice Graphs: Practice sketching the graphs of inverse trigonometric functions. This will help in visualizing their behavior and solving graphical questions efficiently.
Focus on Properties: Familiarise yourself with important properties like $\sin^{-1}(-x) = -\sin^{-1}(x)$ and $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$. These are frequently used in proofs and simplifications.
Work on Conceptual Clarity: Avoid rote memorisation. Instead, focus on understanding why properties and formulas work through examples and derivations.
Conclusion
Vedantu's online learning platform provides important questions for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions. These questions have been curated by subject matter experts to help students revise the chapter thoroughly. These questions cover all the important topics of the chapter, such as the concept of inverse trigonometric functions and their properties, and help students to prepare well for their exams. The questions are designed to test the students' understanding of the concepts and their problem-solving skills. Students can practice these questions to improve their performance in the exams and gain a better understanding of the chapter.
Related Study Materials for Class 12 Maths Chapter 2 Inverse Trigonometric Functions
S. No | Study Materials for Class 12 Maths Chapter 2 Inverse Trigonometric Functions |
1. | CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions Solutions |
2. | CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions Notes |
3. | CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions NCERT Exemplar |
CBSE Class 12 Maths Chapter-wise Important Questions
CBSE Class 12 Maths Chapter-wise Important Questions and Answers cover topics from all 13 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.
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FAQs on CBSE Important Questions for Class 12 Maths Inverse Trigonometric Functions - 2025-26
1. What is the typical mark weightage for Inverse Trigonometric Functions in the Class 12 Maths board exam for the 2025-26 session?
Based on CBSE trends, Chapter 2, Inverse Trigonometric Functions, typically carries a weightage of 5 to 8 marks. Students can expect a combination of question types, including 1-mark MCQs testing principal values, a 3-mark question on proving an identity, and potentially a 4 or 5-mark HOTS question.
2. Which types of questions are most expected for 3 and 5 marks from this chapter in the 2025-26 CBSE board exam?
For higher marks, the most frequently asked questions involve:
- Proving Identities: Questions that require you to prove complex trigonometric identities using standard formulas like those for $2\tan^{-1}x$ or sum and difference formulas.
- Solving Equations: Problems where you must find the value of 'x' by simplifying intricate inverse trigonometric expressions.
- HOTS Questions: Application-based problems that combine multiple concepts, often requiring simplification before applying a known property to find a solution.
3. How are HOTS (Higher Order Thinking Skills) questions structured in Inverse Trigonometric Functions, and how can I score full marks?
HOTS questions in this chapter typically involve simplifying non-standard expressions or solving equations with subtle domain and range constraints. To score full marks, you must:
- Clearly state the properties and identities you are using at each step.
- Show the transformation from the complex form to a simpler one methodically.
- Explicitly mention and verify that values used or obtained lie within the principal value branch of the respective functions.
- Provide a neat, conclusive final answer.
4. Why is a strong understanding of domain and principal value branch so important for scoring well in this chapter?
Understanding the domain and principal value branch is critical because it forms the very definition of an inverse trigonometric function. In the CBSE board exam, evaluators specifically check if your final answers and intermediate steps adhere to these official ranges (e.g., $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for $\sin^{-1}x$). Ignoring these rules is a common reason students lose marks, even if their algebraic manipulation is correct.
5. How does the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ help in solving important board questions?
This identity is a powerful tool for simplification in board exams. Its main importance lies in:
- Inter-conversion: It allows you to convert $\sin^{-1}x$ to $\cos^{-1}x$ and vice-versa, which is crucial for solving equations or proofs where terms need to be in a uniform format.
- Simplification: It helps quickly simplify expressions in 1-mark MCQs and is often a key step in many 3-mark proofs, reducing the complexity of the problem significantly.
6. What are the most common conceptual traps students fall into while solving Inverse Trigonometric Functions questions in board exams?
The most common traps include:
- Ignoring the principal value branch and writing a general value as the final answer.
- Applying formulas like $\tan^{-1}x + \tan^{-1}y$ without checking the required condition (e.g., $xy < 1$).
- Forgetting to check if the final solution for 'x' in an equation lies within the valid domain of the original functions.
- Incorrectly simplifying expressions like $\sin^{-1}(\sin x)$ to $x$ when $x$ is outside the principal value range.
7. What is the best strategy for solving proof-based questions from this chapter to maximise marks?
To maximise marks in proofs, follow a systematic approach as per the CBSE pattern:
- Start with the more complex side (LHS or RHS) of the identity to be proven.
- Clearly state the formula or property you are applying at each step. For example, write, "Using the identity $2\tan^{-1}a = \tan^{-1}(\frac{2a}{1-a^2})$...".
- Show all algebraic simplifications clearly.
- Ensure every step logically follows from the previous one, paying close attention to domain and range conditions.
- Conclude by stating LHS = RHS, hence proved.
8. How does the CBSE marking scheme award marks for a 5-mark question on Inverse Trigonometric Functions?
For a 5-mark question in the 2025-26 exam, marks are typically distributed as follows:
- 1 Mark: For correctly identifying and stating the primary formula or property needed.
- 2-3 Marks: For correct substitution, step-by-step algebraic manipulation, and simplification.
- 1 Mark: For verifying that the values adhere to the principal value branch or domain constraints.
- 1 Mark: For the final, accurate answer. Partial marks are often given for a correct method with a minor calculation error.
9. In an exam, if I solve an equation and get two possible answers, how do I correctly present the solution to get full marks?
To secure full marks, you must demonstrate a complete understanding. First, present both potential solutions you found. Then, you must verify each solution by substituting it back into the original equation or checking it against the domain of the functions involved. Clearly state which solution is valid and provide a brief justification for rejecting the extraneous solution (e.g., "This value is rejected as it does not lie in the domain [-1, 1] of $\cos^{-1}x$").
10. Why is it necessary to check conditions like $xy < 1$ for the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}(\frac{x+y}{1-xy})$ in a board exam?
Checking the condition $xy < 1$ is crucial because the standard identity is only valid within a specific range, which ensures the output lies in the principal value branch of $\tan^{-1}$, i.e., $(-\frac{\pi}{2}, \frac{\pi}{2})$. The CBSE marking scheme for the 2025-26 exam awards marks for conceptual correctness, not just formula application. Failing to verify this condition can lead to a deduction of marks, as it shows an incomplete understanding of the property's limitations.
11. What are the most effective practice habits to score full marks in the Inverse Trigonometric Functions chapter?
To master this chapter for the board exam, you should:
- Create a formula sheet with all identities, their conditions, and the domain/range table for all six inverse functions.
- Solve a variety of proof-based questions from NCERT and past CBSE papers.
- Practise writing answers exactly as per the board pattern, mentioning formulas and justifying each step.
- Pay special attention to questions involving principal value branches, as they are a frequent source of error.





