Class 12 Maths Chapter 2 Miscellaneous Exercise Solutions

\( \operatorname{Cos}^{-1}\left(\cos \frac{13 \pi}{6}\right) \)
(For co\({s}^{-1}\) \( (\cos x) \) 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. \( [0, \pi]) \)
So here, \( \frac{13 \pi}{6} \notin[0, \pi] \)
Now, \( \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \) can be written as,
\( \operatorname{Cos}^{-1}\left(\cos \frac{13 \pi}{6}\right) \)
\( =\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right] \)
\( =\cos ^{-1}\left(\cos \frac{\pi}{6}\right) \) where \( \frac{\pi}{6} \in[0, \pi] \)
\( =\frac{\pi}{6} \)
Hence, \( \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\frac{\pi}{6} \)

\(
\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)
\)
(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. \( \left.\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right) \)
So here, \( \frac{7 \pi}{6} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Now, \( \tan ^{-1}\left(\tan \frac{7 \pi}{6}\right) \) can be written as,
\( =\tan ^{-1}\left[\tan \left(\pi+\frac{\pi}{6}\right)\right] \)
\( =\tan ^{-1}\left(\tan \frac{\pi}{6}\right) \)
where, \( \frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
[since, \( \tan (\pi+x)=\tan x] \)
\( =\frac{\pi}{6} \)
Hence, \( \tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\frac{\pi}{6} \)

Taking LHS
Let
\( \sin ^{-1} \frac{3}{5}=x \)
Then,
\( \operatorname{Sin} x=\frac{3}{5} \)
Therefore,
\( \tan x=\frac{3}{\sqrt{5^{2}-3^{2}}}=\frac{3}{\sqrt{25-9}} \)
\( \therefore \tan x=\frac{3}{4}=x=\tan ^{-1} \frac{3}{4} \)
\( =\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4} \ldots (1)\)
and
\( 2 \sin ^{-1} \frac{3}{5}=2 \tan ^{-1} \frac{3}{4} \)
Now, we know
\( 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \)
\( =2 \tan ^{-1} \frac{3}{4}=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right) \)
\( =\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right) \)
\( =\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right) \)
\( =\tan ^{-1} \frac{24}{7} \)
As, LHS = RHS
Hence Proved!

\( \operatorname{Sin}^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36} \)
\( \operatorname{Sin}^{-1}\left(\frac{8}{17}\right)=x \)
\( \operatorname{Sin} x=\frac{8}{17} \)
\( \operatorname{Cos} x=\sqrt{1-\left(\frac{8}{17}\right)^{2}} \)
\( =\sqrt{\frac{225}{289}} \)
\( =\frac{15}{17} \)
\( \therefore \tan x=\frac{8}{15}=x=\tan ^{-1} \frac{8}{15} \)
\( =\sin ^{-1} \frac{8}{17}=\tan ^{-1} \frac{8}{15} \ldots (1) \)
Let \( \sin -1 \frac{3}{5}=y \)
Then, \( \sin y=\frac{3}{5} \)
\( =\cos y=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5} \)
\( =\tan y=\frac{3}{4}=y=\tan ^{-1} \frac{3}{4} \)
\( =\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4} \ldots (2) \)
Now,
L.H.S. \( =\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5} \)
\( =\tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{3}{4} \) putting the value from equation (1) and (2)
\( =\tan ^{-1} \frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}\left[\right. \) since, tan\( \left.{ }^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1-x y}\right] \)
\( =\tan ^{-1} \frac{32+45}{60-24} \)
\( =\tan ^{-1} \frac{77}{36} \)
\( = \) R.H.S
Hence proved.

\( \operatorname{Cos}^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65} \)
Let \( \cos ^{-1} \frac{4}{5}=x \)
Then, \( \cos x=\frac{4}{5} \)
\( =\sin x=\sqrt{1-\left(\frac{4}{5}\right)^{2}}=\frac{3}{5} \)
\( =\tan x=\frac{3}{4}=x=\tan ^{-1} \frac{3}{4} \)
\( =\cos ^{-1} \frac{4}{3}=\tan ^{-1} \frac{3}{4} \ldots \) (1)
Let, \( \cos ^{-1} \frac{12}{13}=y \)
Then \( \cos y=\frac{12}{13} \)
\( =\sin y=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\frac{5}{13} \)
\( =\tan y=\frac{5}{12}=y=\tan ^{-1} \frac{5}{12} \)
\( =\cos ^{-1} \frac{12}{13}=\tan ^{-1} \frac{5}{12} \ldots (2)\)
Let, \( \cos ^{-1} \frac{33}{65}=z \)
Then, \( \cos z=\frac{33}{65} \)
\( =\sin z=\sqrt{1-\left(\frac{33}{65}\right)^{2}}=\frac{56}{65} \)
\( \tan z=\frac{56}{33}=y=\tan -1 \frac{56}{33} \)
\( =\cos -1 \frac{33}{65}=\tan -1 \frac{56}{33} \ldots (3)\)
Now,
L.H.S. \( =\cos -1 \frac{4}{5}+\cos ^{-1} \frac{12}{13} \)
\( =\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12} \) putting the value from the equation \( (1) \) and \( (2) \)
\( =\tan ^{-1} \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \times \frac{5}{12}}\left[\right. \) since, tan \( \left.{ }^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1-x y}\right] \)
\( =\tan ^{-1} \frac{36+20}{48-15} \)
\( =\tan ^{-1} \frac{56}{33} \ldots \) by equation \( (3) \)
\( = \) R.H.S.
Hence, proved.

\(
\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}
\)
We can also solve this problem by using the identity \( \operatorname{Sin}(A+B)=\sin A \) \( \operatorname{Cos} \mathrm{B}+\cos \mathrm{A} \operatorname{Sin} \mathrm{B} \)
Let, \( \sin -1 \frac{3}{5}=A \) and \( \cos -1 \frac{12}{13}=B \)
So,
\( \sin A=\frac {3} {5} \) and \( \cos B= \frac{12}{13} \) Therefore, \( \cos A=\frac{4} {5} \) and \( \sin B=\frac {5} {13} \)
As R.H.S. is sin-1 we will use \( \sin (\mathrm{A}+\mathrm{B}) \)
\(
\begin{array}{l}
=\sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \operatorname{Cos} \mathrm{B}+\cos \mathrm{A} \operatorname{Sin} \mathrm{B} \\
=\frac{3}{5} \times \frac{12}{13}+\frac{4}{5} \times \frac{5}{13}=\frac{36}{65}+\frac{20}{65} \\
=\frac{56}{65}
\end{array}
\)
Thus, \( \mathrm{A}+\mathrm{B}=\sin ^{-1} \frac{56}{65} \)
\( = \) R.H.S.
Hence Proved.

\(
\tan ^{-1} \frac{63}{16}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}
\)
let, \( \sin ^{-1} \frac{5}{13}=x \) then, \( \sin x=\frac{5}{13} \)
\( =\cos x=\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\frac{12}{13} \)
\( =\tan x=\frac{5}{12}=x=\tan ^{-1} \frac{5}{12} \)
\( =\sin ^{-1} \frac{5}{13}=\tan ^{-1} \frac{5}{12} \ldots (1)\)
Let, \( \cos ^{-1} \frac{3}{5}=y \). then, \( \cos y=\frac{3}{5} \)
\( =\sin y=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5} \)
\( =\tan y=\frac{4}{3}=y=\tan ^{-1} \frac{4}{3} \)
\( =\cos ^{-1} \frac{3}{5}=\tan ^{-1} \frac{4}{3} \ldots (2)\)
Now,
R.H.S. \( =\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5} \)
\( =\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{4}{3} \) putting the value from equation (1) and (2)
\( =\tan ^{-1} \frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}\left(\right. \) since, tan \( \left.{ }^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1-x y}\right) \)
\( =\tan ^{-1} \frac{15+48}{36-20} \)
\( =\tan ^{-1} \frac{63}{16} \)
\( = \) L.H.S.
Hence, proved.

L.H.S. \( =\tan -1 \frac{1}{5}+\tan -1 \frac{1}{7}+\tan -1 \frac{1}{3}+\tan -1 \frac{1}{8} \)
\( =\tan -1 \frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5} \times \frac{1}{7}}+\tan -1 \frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3} \times \frac{1}{38}} \quad\left[\text{Since}, {tan}^{-1} x-\tan ^{-1} y= \tan ^{-1} \frac{x-y}{1-x y}\right] \)
\( =\tan -1 \frac{7+5}{35-1}+\tan -1 \frac{8+3}{24-1} \)
\( =\tan ^{-1} \frac{12}{34}+\tan -1 \frac{11}{23} \)
\( =\tan ^{-1} \frac{6}{17}+\tan -1 \frac{11}{23} \)
\( =\tan ^{-1} \frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17} \times \frac{11}{23}}\left[\right. \) since, \( \left.\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1-x y}\right] \)
\( =\tan ^{-1} \frac{138+187}{391-66} \)
\( =\tan ^{-1} \frac{325}{325}=\tan ^{-1}=\frac{\pi}{4} \)
\( = \) R.H.S.
Hence, proved.

\( \tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right) \)
Let \( x=\tan 2 \theta \).Then, \( \sqrt{x}=\tan \theta \)
\( =\theta=\tan ^{-1} \sqrt{x} \)
\( \therefore \frac{1-x}{1+x}=\frac{1-\tan ^{2}}{1+\tan ^{2}}=\cos 2 \theta \)
So now putting the value, we get,
R.H.S. \( =\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right) \)
\( =\frac{1}{2} \cos ^{-1}(\cos 2 \theta)=\frac{1}{2} \times 2 \theta=\tan ^{-1} \sqrt{x} \)
\( = \) L.H.S.

\( \operatorname{Cot}^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) \)
Consider, \( \left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) \)
On Rationalizing, we get,
\( =\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1+\sin x})^{2}} \)
\( =\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1+\sin x)(1-\sin x)}}{1+\sin x-1+\sin x} \)
\( =\frac{2\left(1+\sqrt{1-\sin ^{2} x}\right)}{2 \sin x}=\frac{1+\cos x}{\sin x}=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin{\frac{x}{2}} \cos \frac{x}{2}}=\cot \frac{x}{2} \)
Now,
L.H.S. \( =\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) \)
\(
\begin{array}{l}
=\cot ^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2} \\
=\text { R.H.S. }
\end{array}
\)
Hence Proved

\(
\tan -1\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos -1 x
\)
Let \( x=\cos 2 \theta \) so that \( \theta=\frac{1}{2} \cos -1 x \)
Now,
\(
\begin{array}{l}
\text { L.H.S. }=\tan -1\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) \\
=\tan -1\left(\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right) \\
=\tan ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}\right) \\
=\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}\right) \\
=\tan ^{-1}\left(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right) \\
=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \\
=\tan ^{-1} 1-\tan { }^{-1}(\tan \theta)=\frac{\pi}{4}-\theta \\
=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x
\end{array}
\)
\( = \) R.H.S.
Hence Proved.

\(
\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}
\)
Now, L.H.S. \( \frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3} \)
\(\begin{array}{l}
=\frac{9}{4}\left(\frac{\pi}{2}-si{n}^{-1} \frac{1}{3}\right) \\
=\frac{9}{4} \times \cos ^{-1} \frac{1}{3} \ldots \text { (1) }
\end{array}\)
Now, Let
\( \operatorname{Cos}^{-1} \frac{1}{3}=x \). \( \text{then }\cos x=\frac{1}{3}=\sin x=\sqrt{1-\left(\frac{1}{3}\right)^{3}}=\frac{2 \sqrt{2}}{3} \)
\( \therefore x=\sin ^{-1} \frac{2 \sqrt{2}}{3}=\cos ^{-1} \frac{1}{3}=\sin ^{-1} \frac{2 \sqrt{2}}{3} \)
L.H.S. \( =\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3} \)
\( = \) R.H.S.
Hence Proved

\(
\begin{array}{l}
2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x) \\
=\tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)=\tan ^{-1}(2 \operatorname{cosec} x) \\
=\frac{2 \cos x}{1-\cos ^{2} x}=2 \operatorname{cosec} x \\
=\frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x}=\cos x=\sin x \\
=\tan x=1
\end{array}
\)
Hence, \( x=\frac{\pi}{4} \)

\(
\tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x\)
\( \text{As we know,} \tan ^{-1}(x)-\tan ^{-1}(y)=\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \)
\(
=\tan ^{-1} 1-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x
\)
We know, \( \tan \frac{\pi}{4}=1 \)
So, \( \tan ^{-1}(1)=\frac{\pi}{4} \)
\(
\begin{array}{l}
=\frac{\pi}{4}-\tan ^{-1}(x)=\frac{1}{2} \tan ^{-1}(x) \\
=\frac{\pi}{4}=\frac{3}{2} \tan ^{-1} x \\
=\tan ^{-1} x=\frac{\pi}{6} \\
=x=\tan \frac{\pi}{6}
\end{array}
\)
Hence,
\(
x=\frac{1}{\sqrt{3}}
\)

\( \text{Let}\tan -1 x=y \), \( \text{then }\tan y=x=\sin y=\frac{x}{\sqrt{1+x^{2}}} \)
\( \therefore y=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right) \)
\( =\tan ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right) \)
\( =\sin \left(\tan ^{-1} x\right)=\sin \left(\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)\right) \)
\( =\frac{x}{\sqrt{1+x^{2}}} \)

\( \operatorname{Sin}^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2} \)
Now we will put Now, we will put \( x=\sin y \) in the given equation, and we get,
\(\operatorname{Sin}^{-1}(1-\sin y)-2 \sin ^{-1} \sin y=\frac{\pi}{2}\)
\(=\sin ^{-1}(1-\sin y)-2 y=\frac{\pi}{2} \pi\)
\(=\sin ^{-1}(1-\sin y)=\frac{\pi}{2}+2 y\)
\(=1-\sin y=\sin \left(\frac{\pi}{2}+2 y\right)\)
\(\Rightarrow 1-\sin y=\cos 2 y \text { as } \sin \left(\frac{\pi}{2}+x\right)=\cos x\)
\(\Rightarrow 1-\cos 2 y=\sin y\)
\(\Rightarrow 2 \sin 2 y=\sin y\)
\(\Rightarrow \sin y(2 \sin y-1)=0\)
\(\Rightarrow \sin y=0 \text { or } \frac{ 1 }{ 2 }\)
\(\therefore x=0 \text { or } \frac{ 1 }{ 2 }\)
Now, if we put then we will see that,
L.H.S. \( =\sin ^{-1}\left(\frac{1}{2}\right)=-2 \sin ^{-1} \frac{1}{2} \)
\(=-\sin ^{-1} \frac{1}{2}\)
\( =-\frac{\pi}{6} \neq \frac{\pi}{2} \neq \) R.H.S
Hence, \( x=\frac{1}{2} \) is not the solution of the given equation.
Thus, \( x=0 \)

\(
\tan ^{-1} \frac{x}{y}-\tan ^{-1} \frac{x-y}{x+y}\)
\(=\tan ^{-1}\left[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1-\frac{x}{y} \cdot \frac{x-y}{x+y}}\right]\)
\(=\tan ^{-1}\left[\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right]\)
\(=\tan ^{-1}\left(\frac{x^{2}+x y-x y+y^{2}}{x y+y^{2}+x^{2}-x y}\right)\)
\(=\tan ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}+y^{2}}\right)=\tan ^{-1} 1=\frac{\pi}{4}
\)