Exercise 2.2 class 12 maths ncert solutions

Let \( x=\sin \theta \) then \( \sin ^{-1} x=\theta \)
We have,
\(
\text { R.H.S }=\sin ^{-1}\left(3 x-4 x^{3}\right)^{3}=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right)
\)
Now, we know that,
\(
\sin 3 x=3 \sin x-4 \sin 3 x
\)
Therefore,
\(
=\sin ^{-1}(\sin (3 \theta))\)
\(=3 \theta\)
\(=3 \sin ^{-1} x\)
\(=\text { L.H.S }
\)
Hence Proved

Let \( x=\cos \theta \)
Then, \( \operatorname{Cos}^{-1}=\theta \)
Now, R.H.S. \( =\cos ^{-1}\left(4 x^{3}-3 x\right) \)
\(
=\cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right)\)
\(=\cos ^{-1}\left(\cos ^{3} \theta\right)\)
\(=3 \theta\)
\(=3 \cos ^{-1} x\)
\(=\text { L.H.S. }
\)
Hence Proved

L.H.S. \( \tan -1 \frac{2}{11}+\tan -1 \frac{7}{24} \)
\(
=\tan ^{-1} \frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11} \cdot \frac{7}{24}} \quad\left[\therefore \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right]\)
\(=\tan ^{-1} \frac{\frac{48+77}{11 \times 24}}{\frac{11 \times 24-14}{11 \times 24}}\)
\(=\tan ^{-1} \frac{48+77}{264-14}\)
\(=\tan ^{-1} \frac{125}{250}\)
\(=\tan ^{-1} \frac{1}{2}\)
\(=\text { R.H.S. }\)
Hence Proved.

L.H.S. \( =2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7} \)
\( =\tan ^{-1} \frac{2 \cdot \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}+\tan ^{-1} \frac{1}{7} \)
\( =\tan ^{-1} \frac{1}{\frac{3}{4}}+\tan ^{-1} \frac{1}{7} \)
\( =\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7} \)
\( =\tan ^{-1} \frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \cdot \frac{1}{7}}\left[\right. \) since, \( \left. \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right] \)
\( =\tan ^{-1} \frac{\frac{28+3}{21}}{\frac{21-4}{21}}\)
\( =\tan ^{-1} \frac{31}{17} \)
\( = \) R.H.S.
Hence Proved.

\(
\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}, x \neq 0
\)
Now, Put \( x=\tan \theta \Rightarrow \theta=\tan ^{-1} x \)
Therefore, \( \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}=\tan ^{-1}\left(\frac{\sqrt{1+\tan \theta^{2}}-1}{\tan \theta}\right) \)
\( =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \)
\( =\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \)
\( =\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin{\frac{\theta}{2}}\cos \frac{\theta}{2}}\right) \)
\( =\tan ^{-1}\left(\tan \frac{\theta}{2}\right) \)
\( =\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x \)
Therefore, \( \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}=\frac{1}{2} \tan ^{-1} x \)

\(
\tan ^{-1} \frac{1}{\sqrt{x^{2}-1}},|x| > 1
\)
Let us take,
\(
x=\operatorname{cosec} \theta=\theta=\operatorname{cosec}^{-1} x
\)
[We have done this substitution on the bases of identity \( \sec 2 \theta-1= \) \( \tan 2 \theta] \)
Therefore, \( \tan ^{-1} \frac{1}{\sqrt{x^{2}-1}}=\tan ^{-1} \frac{1}{\sqrt{\operatorname{cosec}^{2} \theta-1}} \)
Now we know that, \( \operatorname{cosec} 2 \theta-1=\cot 2 \theta \)
Therefore,
\( =\tan ^{-1} \frac{1}{\cot \theta}=\tan ^{-1}(\tan \theta)=\theta=\operatorname{cosec}^{-1} x \)

\( \tan ^{-1}=\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right) x < \pi \)
\( =\tan ^{-1}\left(\sqrt{\frac{2 \sin^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}\right) \)
\( =\tan ^{-1}\left(\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right) \)
\( =\tan ^{-1}\left(\tan \frac{x}{2}\right) \)
\( =\frac{x}{2} \)
Hence, \( \tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\frac{x}{2} \)

\(
\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)
\)
Dividing by \( \cos x \),
\( =\tan ^{-1}\left(\frac{1-\frac{s i n x}{\cos x}}{1+\frac{s i n x}{\cos x}}\right) \)
\( =\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right),\left[\therefore \frac{\sin x}{\cos x}=\tan x\right] \)
\( =\tan ^{-1}-\tan ^{-1}(\tan x)\left[\therefore \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right] \)
As we know \( \tan (\frac {\pi}{4})=1 \)
\( =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}\right)\right)+\tan ^{-1}(\tan x) \)
\( =\frac{\pi}{4}-x \)
Hence, \( \tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi}{4}-x \).

\(
\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}
\)
We will solve this problem on the bases of the identity \( 1-\sin 2 \theta=\cos 2 \theta \)
So, for \( \mathrm{a}^{2}-x^{2} \), we can substitute \( x=\mathrm{a} \sin \theta \) or \( x=\mathrm{a} \cos \theta \)
Now, let us put \( x=\mathrm{a} \sin \theta \)
\(
=\frac{x}{a}=\sin \theta\)
\(=\theta=\sin ^{-1}\left(\frac{x}{a}\right)
\)
Therefore,
\(\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}=\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right)=\tan ^{-1}\left(\frac{a \sin \theta}{a \cos \theta}\right)\)
\(=\tan ^{-1}(\tan \theta)=\theta=\sin ^{-1}\left(\frac{x}{a}\right)
\)
Hence, \( \tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1}\left(\frac{x}{a}\right) \)

\(
\tan ^{-1}\left\{\frac{3 a^{3} x-x^{3}}{a^{3}-3 a x^{2}}\right\}
\)
Put \( x=\mathrm{a} \tan \theta \Rightarrow \frac{x}{a}=\tan \theta=\tan ^{-1} \frac{x}{a} \)
Now,
\(
\tan ^{-1}\left(\frac{3 a^{3} x-x^{3}}{a^{3}-3 a x^{2}}\right)=\tan ^{-1}\left(\frac{3 a^{3} \cdot a \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot a^{2} \tan ^{2} \theta}\right)\)
\(=\tan ^{-1}\left(\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}\right)\)
\(=\tan ^{-1}(\tan 3 \theta)\left[\therefore \frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}=\tan 3 \theta\right]\)
\(=3 \theta\)
\(=3 \tan ^{-1} \frac{x}{a}
\)

\(
\tan ^{-1}\left[2 \cos \left(2 \operatorname{si} \bar{n}^{1} \frac{1}{2}\right)\right]
\)
We will solve the inner bracket first.
So, we will first find the principal value of \( \sin ^{-1} \frac{1}{2} \)
We know that, \( \sin ^{-1} \frac{1}{2}=\frac{\pi}{6} \)
Therefore,
\(
\tan ^{-1}\left[2 \cos \left(2 \operatorname{si} \bar{n}^{1} \frac{1}{2}\right)\right]=\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]\)
\(=\tan ^{-1}\left[2 \cos \left(\frac{\pi}{3}\right)\right]\)
\(=\tan ^{-1}\left[2 \times \frac{1}{2}\right]\left[\text { since, cos }\left(\frac{\pi}{3}\right)=\frac{1}{2}\right]\)
\(=\tan ^{-1} 1\)
\(=\frac{ \pi }{ 4 }
\)
Hence,
The value of \( \tan ^{-1}\left[2 \cos \left(2 \operatorname{si} {n}^{-1} \frac{1}{2}\right)\right]=\frac{\pi}{4} \)

\(
\cot \left(\tan ^{-1} a+\cot ^{-1} a\right)\)
\(=\cot \left(\frac{\pi}{2}\right),\left[\therefore \tan ^{-1} x+\cot ^{-1} y=\frac{\pi}{2}\right]\)
\(=0
\)
Hence, the value of \( \cot \left(\tan ^{-1} a+\cot ^{-1} a\right)=0 \)

\(
\tan \frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right]
\)
We will solve this problem by expressing \( \sin 2 \theta \) and \( \cos 2 \theta \) in terms of \( \tan \theta \)
Now let us put, \( x=\tan \theta \). Then we will have,
\(
{l}
\theta=\tan ^{-1} x \\
\therefore \sin ^{-1} \frac{2 x}{1+x^{2}}=\sin ^{-1} \frac{2 \tan \theta}{1+\tan ^{2} \theta}=\sin ^{-1}(\sin 2 \theta)=2 \theta=2 \tan ^{-1} x
\)
Now again, Let's put, \( y=\tan \emptyset \). Then we will have,
\( \phi=\tan -1 y \)
\(
\therefore \cos ^{-1} \frac{1-y^{2}}{1+y^{2}}=\sin ^{-1}\left(\frac{1-\tan ^{2} \phi}{1+\tan ^{2} \phi}\right)=\cos ^{-1}(\cos 2 \phi)=2 \phi=2 \tan ^{-1} y
\)
Now,
\(
\begin{array}{l}
\tan \frac{1}{2}\left[\sin {n}^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right] \\
=\tan \frac{1}{2}\left[2 \tan ^{-1} x+2 \tan ^{-1} y\right] \\
=\tan \left[\tan ^{-1} x+\tan ^{-1} y\right] \\
=\tan \left[\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right] \\
=\frac{x+y}{1-x y}
\end{array}
\)
Hence, the value of
\( \tan \frac{1}{2}\left[\sin^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right]=\frac{x+y}{1-x y} \)

\(
\begin{array}{l}
\operatorname{Sin }\left(\ si{n}^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1 \\
=\sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\sin -11 \\
=\sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2},\left[\text { since, } si{n}^{-1} 1=\frac{\pi}{2}\right] \\
=\sin ^{-1} \frac{1}{5}=\frac{\pi}{2}-\cos ^{-1} x \\
=\sin ^{-1} \frac{1}{5}=\sin ^{-1} x,\left[\text { since, } si {n}^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]
\end{array}
\)
On comparing the co-efficient on both sides we get,
\(
=x=\frac{1}{5}
\)

\(
\operatorname{Sin}^{-1}\left(\sin _{3}^{2 \pi}\right)
\)
(For \( \sin ^{-1}(\sin x) \) type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
So here, \( \frac{2 \pi}{2} otin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Now, \( \sin ^{-1}\left(\sin \frac{2\pi}{3}\right) \) can be written as,
\( \sin ^{-1}\left(\sin \frac{2\pi}{3}\right) \)
\( =\sin ^{-1}\left(\sin \pi-\frac{\pi}{3}\right) \)
\( =\sin ^{-1}\left(\sin \frac{\pi}{3}\right) \) where \( \frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
\( =\frac{\pi}{3} \)
Hence, \( \sin ^{-1}\left(\sin \frac{2\pi}{3}\right)=\frac{\pi}{3} \).

\(
\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)
\)
(For \( \tan ^{-1}(\tan x) \) type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range. \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) )
So here, \( \frac{3 \pi}{4} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Now, \( \tan ^{-1}\left(\tan \frac{3 \pi}{4}\right) \) can be written as,
\(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\)
\(=\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{4}\right)\right]\)
\( =-\tan -1\left(\tan \frac{\pi}{4}\right) \)
where \( -\frac{\pi}{4} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \),
[since, \( \tan (\pi-x)=-\tan x \)
\( =-\frac{\pi}{4} \)
Hence, \( \tan -1\left(\tan \frac{3 \pi}{4}\right)=-\frac{\pi}{4} \)

Let \( \sin ^{-1}\left(\frac{3}{5}\right)=y \) so \( \sin y=\frac{3}{5} \) and \( y \in\left(0, \frac{\pi}{2}\right) \), so all ratio of y are positive and
Hence, \( \cos y=\frac{4}{5} \) and \( \tan y=\frac{3}{4} \), so \( ta{n}^{-1}\left(\frac{3}{4}\right)=y \)
Also,
\( \operatorname{Cot}^{-1}\left(\frac{3}{2}\right)=\tan ^{-1} \frac{2}{3} \) as \( \cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\right) \)
So, \( \tan \left({si}{n}^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right) \)
\( =\tan \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right) \)
\( =\tan \left(\tan ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right) \)
\( =\tan \left(\tan ^{-1} \frac{17}{6}\right)=\frac{17}{6} \)
Hence, \( \tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)=\frac{17}{6} \)

\( \operatorname{Cos}^{-1}\left(\cos \frac{7 \pi}{6}\right) \)
(For \( \cos ^{-1}(\cos x) \) type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range. \( \left.\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right) \)
So here,
\( \frac{7 \pi}{6} \notin[0, \pi] \)
Now, \( \cos ^{-1}\left(\cos \frac{7 \pi}{67}\right) \) can be written as,
\( \operatorname{Cos}^{-1}\left(\cos \frac{7 \pi}{6}\right) \)
\( =\cos ^{-1}\left[\cos \left(\pi+\frac{\pi}{6}\right)\right] \)
\( =-\cos ^{-1}\left(\cos \frac{\pi}{6}\right) \)
where \( -\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \)
[since, \( \cos (\pi+x)=-\cos x] \)
\( =\pi-\cos ^{-1}\left(\cos \frac{\pi}{6}\right) \) as \( \cos -1(-x)=\pi-\cos -1 \)
\( =\pi-\frac{\pi}{6}=\frac{5 \pi}{6} \)
Hence, \( \cos \left(\cos \frac{7 \pi}{6}\right)=\frac{5 \pi}{6} \).

\( \operatorname{Sin}^{-1}\left(-\frac{1}{2}\right)=\sin ^{-1}\left(\frac{1}{2}\right) \) as \( \sin ^{-1}(-x)=\sin ^{-1} x \)
\( =-\frac{\pi}{6} \) as \( \sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \)
We all know that the principal value branch of \( \sin ^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
\( \therefore \) \( si{n}^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6} \)
Therefore,\( \sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)=\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin \left(\frac{3 \pi}{6}\right)=\sin \left(\frac{\pi}{2}\right)=1 \) Hence, the value of \( \sin \left(\frac{\pi}{3}-\right. \) \( \left. si{n}^{-1}\left(-\frac{1}{2}\right)\right) \)

\(
\tan ^{-1} \sqrt{3-} \cot ^{-1}(-\sqrt{3})\)
\(=\tan ^{-1} \sqrt{3}-\left(\pi-\cot ^{-1} \sqrt{3}\right)\)
\(=\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}-\pi\)
\(=\frac{\pi}{2}-\pi\)
\(=-\frac{\pi}{2}
\)