Practice Problems for CBSE Class 12 Maths Chapter 2: Inverse Trigonometric Functions with FREE PDF
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.
The free PDF also contains solutions toClass 12 Maths Important Questions which are developed as practice exercises for students so that they can improve their subject knowledge. Students can download the free PDF available on Vedantu to prepare for their exams.
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.
S. No | Chapter-wise Important Questions for Class 12 Maths |
1 | |
2 | |
3 | |
4 | Chapter 5 - Continuity and Differentiability Important Questions |
5 | |
6 | |
7 | |
8 | |
9 | |
10 | |
11 | |
12 |
Additional Study Materials for Class 12 Maths
S.No | Study Materials for Class 12 Maths |
1 | |
2 | |
3 | |
4 | |
5 | |
6 | |
7 | |
8 |
FAQs on Class 12 Important Questions: CBSE Maths Chapter 2 Inverse Trigonometric Functions 2024-25
1. What are the most expected 3-mark and 5-mark questions from the CBSE Class 12 Maths Inverse Trigonometric Functions chapter for the 2025–26 board exam?
For the 2025–26 board exam, questions often involve proving identities using inverse trigonometric functions, such as transforming sums or differences into single inverse expressions, or applying composite and double-angle formulas. Other high-weightage questions require solving equations for unknowns using properties of principal branches. Sample questions include deriving properties like $2\tan^{-1}x = \sin^{-1}(\frac{2x}{1+x^2})$, evaluating expressions like $\tan^{-1}x + \tan^{-1}y = \tan^{-1}(\frac{x+y}{1-xy})$, and applying inverse identities in HOTS questions.
2. How is marking for long-answer questions (4–5 marks) structured in CBSE Class 12 Inverse Trigonometric Functions?
Long-answer questions carry marks for each logical step: 1 mark for correct formula identification, 2–3 marks for proper substitution and stepwise manipulation (using identities and domains/ranges), 1 mark for clear justification of principal value choices, and 1 mark for correct final simplification.
- Partial marks are awarded if the method is correct but there is an algebraic slip, as per the CBSE 2025–26 marking scheme.
- Boxing or clearly phrasing your final answer is recommended for full marks.
3. How do you avoid common mistakes when solving board-level inverse trigonometric function questions?
To avoid errors in these questions:
- Always check the domain and range of each function before substituting values.
- Use the principal value branch as per CBSE rules (e.g., for $\sin^{-1}x$, range is $[-\frac{\pi}{2},\frac{\pi}{2}]$).
- Justify each transformation, particularly when using sum/difference identities.
- Strike out extraneous roots with a note explaining why they don't fit the principal branch.
4. Why are questions involving domain and range considered high-yield in this chapter?
Questions about domain and range directly test understanding of function definition, principal branches, and allowed input values. They are easy scoring opportunities since answers are fixed and tables are provided in the CBSE pattern. Including them ensures coverage of foundational knowledge essential for marks both in short and long answers.
5. What is the importance of the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ in solving exam questions?
This identity enables direct conversions between $\sin^{-1}$ and $\cos^{-1}$ answers, simplifying proofs or calculations involving composite expressions. It frequently appears in 1-mark MCQs and multi-step proofs, and using it correctly ensures you remain within the principal value ranges required by CBSE.
6. Which board-level conceptual traps are common in inverse trigonometric function questions?
Common traps include:
- Using the general value instead of the principal value branch required by CBSE.
- Ignoring whether $x$ lies in the correct domain or range for the function.
- Misapplying sum and difference identities by not checking conditions like $xy < 1$ for $\tan^{-1}x + \tan^{-1}y$.
- Assuming all roots of an equation are valid, without eliminating extraneous values that don’t fit the defined branch.
7. How do you efficiently choose which identity to apply for composite or proof-based questions?
Analyze the structure of the expression: If it contains sums or differences of inverse functions, check if addition/subtraction identities fit. For double angles, use $2\tan^{-1}x$ formulas. Always compare the question format against standard CBSE reference sheets and confirm the domain/range of every substitution.
8. What are HOTS (Higher Order Thinking Skills) question types in this chapter, and how should they be solved for full marks?
HOTS questions include:
- Proving or transforming non-standard expressions (like reducing $\sin^{-1}(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2})$) into standard forms.
- Solving equations involving multiple inverse trigonometric terms with hidden domains.
9. How can a student distinguish between a question requiring proof and one needing evaluation in board exams?
Proof-based questions typically ask you to derive or prove a given relationship (e.g., show that two expressions are equivalent for all valid $x$), requiring step-by-step logic and use of formulas. Evaluation questions ask you to compute a function value for a specific $x$ or parameter, focusing more on substitution and simplification within CBSE’s principal branches.
10. What are the most effective strategies for quickly solving composite inverse trigonometric function questions in CBSE board exams?
- Substitute the innermost expression first and check that its output is within the domain of the next function.
- Break complex expressions into standard forms using identities (e.g., express $\tan^{-1}(\tan x)$ as $x$ only if $x$ is in the defined range).
- Always confirm that the final answer lies in the prescribed principal value branch to avoid CBSE marking deductions.
11. If a solution leads to an extraneous root or a value outside the principal range, what must be written in a CBSE exam answer?
State all possible roots as part of your answer, then clearly check and eliminate those not within the principal branch or valid domain, marking the rejected ones with a brief justification. This demonstrates full understanding and receives marks as per CBSE 2025–26 rules.
12. In what ways does the Inverse Trigonometric Functions chapter prepare students for competitive exams like JEE and NEET?
This chapter builds a foundation in function composition, property derivation, and algebraic manipulation—all of which are common in JEE and NEET maths. Many competitive exam questions are modeled after board trends and will reuse identities, proofs, and principal value logic from this syllabus, making practice here directly relevant for further exams.
13. How often do inverse trigonometric function questions appear in CBSE Class 12 board papers and what is their typical mark weightage?
Inverse trigonometric function questions generally account for 6–10 marks in the CBSE Class 12 board exam, typically including at least one short (1–2 marks), one moderate (3-mark), and one long (4–5 mark) question, plus possible inclusion in integration/calculus problems.
14. What is a good approach for tackling proof-based inverse function questions to maximize marks?
Always start by stating the identity or property to be used, break down each algebraic or trigonometric manipulation into steps, justify your use of principal branches and domains, write all key intermediate results, and provide a clearly boxed final answer. This approach fits the recommended CBSE format and secures maximum marking.
15. Which specific practice habits help students score higher in board exam questions from the Inverse Trigonometric Functions chapter?
- Make a formula sheet of key inverse trigonometric identities and principal value ranges.
- Practice applying identities in both directions (expansion and reduction).
- Focus on writing stepwise solutions with explicit domain and range checks.
- Analyze past CBSE board questions to identify frequently repeated proof structures.

















