Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 1

Question 1. Find the value of ${\cos }^{-1}(\cos \frac {13\pi} {6})$

Solution:

We know that $\cos^{-1} (\cos x)=x$
Here, $\frac {13\pi} {6} \notin [0,\pi].$
Now, ${\cos }^{-1}(\cos \frac {13\pi} {6})$ can be written as :
${\cos }^{-1}(\cos \frac {13\pi} {6})={\cos }^{-1}[\cos( 2\pi+\frac {\pi} {6})]$ , where $\frac{\pi} {6} \in [0,\pi].$
Hence, the value of ${\cos }^{-1}(\cos \frac {13\pi} {6})$ = π/6

Question 2. Find the value of $\tan^{-1}(\tan \frac {7\pi}{6})$

Solution:

We know that $\tan^{-1} (\tan x)=x$
Here, $\frac {7\pi}{6} \notin(\frac{-\pi}{2},\frac{\pi}{2})$
Now, $\tan^{-1}(\tan \frac {7\pi}{6})$ can be written as:
$\tan^{-1}(\tan \frac {7\pi}{6})=\tan^{-1}[\tan( 2\pi -\frac {5\pi}{6})]$ $-[\tan(2\pi-x)=-\tan x]$
$\tan^{-1}[-\tan(\frac {5\pi}{6}) ]=\tan^{-1}[\tan(-\frac {5\pi}{6})]$
$=\tan^{-1}[\tan(\pi-\frac {5\pi}{6})]=\tan^{-1}[\tan( \frac {\pi}{6})],$ where $\frac{\pi}{6} \in (\frac{-\pi}{2},\frac{\pi}{2})$
Hence, the value of $\tan^{-1}(\tan\frac{7\pi}{6})$ = π/6

Question 3. Prove $2\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{24}{7}$

Solution:

Let $\sin^{-1} \frac{3}{5}=x$ -(1)
sin x = 3/5
So, $\cos x = \sqrt{1-(\frac{3}{5})^2 }$ = 4/5
tan x = 3/4
Hence, $x=\tan^{-1} \frac{3}{4}$
Now put the value of x from eq(1), we get
$\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{3}{4}$
Now, we have
L.H.S $= 2 \sin^{-1} \frac{3}{5}=2 \tan^{-1} \frac{3}{4}$
= $\tan^{-1}(\frac{2 \times \frac{3}{4}}{1-(\frac{3}{4})^{2}})$ – $[2\tan^{-1} x=\tan^{-1} \frac{2x}{1-x^2}]$
$= \tan^{-1}(\frac{ \frac{3}{2}}{\frac{16-9}{16}})=\tan^{-1} (\frac{3}{2} \times \frac{16}{7})$
$=\tan^{-1} \frac{24}{7}$
Hence, proved.

Question 4. Prove $\sin^{-1} \frac{8}{17}+\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{77}{36}$

Solution:

Let $\sin^{-1} \frac{8}{17}=x$
Then sin x = 8/17
cos x = $\sqrt{1-(\frac{8}{17})^2}=\sqrt \frac{225}{289}$ = 15/17
Therefore, $\tan x=\frac{8}{15}\implies x=\tan^{-1}\frac{8}{15}$
$\sin^{-1} \frac{8}{17}=\tan^{-1} \frac{8}{17}$ -(1)
Now, let $\sin^{-1} \frac{3}{5}=y$
Then, sin y = 3/5
$\cos y=\sqrt{1- (\frac{3}{5})^2}=\sqrt{ (\frac{16}{25})}$ = 4/5
$\therefore \tan y =\frac{3}{4} \implies y=\tan^{-1} \frac{3}{4}$
$\therefore \sin^{-1} \frac{3}{5}=\tan^{-1} \frac{3}{4}$ -(2)
Now, we have:
L.H.S. $=\sin^{-1} \frac{8}{17}+\sin^{-1} \frac{3}{5}$
From equation(1) and (2), we get
= $\tan^{-1} \frac{8}{15}+\tan^{-1} \frac{3}{4}$
= $\tan^{-1} \frac{{\frac{8}{15}+ \frac{3}{4}} }{1-{\frac{8}{15}\times \frac{3}{4}}}$
= $\tan^{-1}(\frac{32+45}{60-24})$ – $[\tan^{-1} x + \tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$
= $\tan^{-1} \frac{77}{36}$
Hence proved

Question 5. Prove $\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$

Solution:

Let $\cos^{-1}\frac{4}{5}= x$
Then, cos x = 4/5
$\sin x = \sqrt {1- (\frac{4}{5})^{2}}$ = 3/5
$\therefore \tan x =\frac{3}{4} \implies x=\tan^{-1} \frac{3}{4}$
$\therefore \cos^{-1} \frac{4}{5}=\tan^{-1} \frac{3}{4}$ -(1)
Now let $\cos^{-1} \frac{12}{13}=x$
Then, cos y = 3/4
$\sin^{-1} y=\frac{5}{13}$
$\therefore\tan y= \frac{5}{12} \implies y=\tan^{-1} \frac{5}{12}$
$\therefore \cos ^{-1} \frac{12}{13}=\tan^{-1} \frac{5}{12}$ -(2)
Let $\cos^{-1} \frac{33}{65}=z$
Then, cos z = 33/65
sin z = 56/65
$\therefore \tan z = \frac{56}{65} \implies z= tan^{-1}\frac{56}{33}$
$\therefore \cos^{-1} \frac{33}{65}= \tan^{-1} \frac{56}{33}$ -(3)
Now, we will prove that :
L.H.S. $=\cos^{-1} \frac{3}{5}\cos^{-1} \frac{12}{13}$
From equation (1) and equation (2)
= $\tan^{-1} \frac{3}{4}+\tan^{-1} \frac{5}{12}$
= $\tan^{-1} \frac{{\frac{3}{4}+ \frac{5}{12}} }{1-{\frac{3}{4}\times \frac{5}{12}}}$ – $[\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$
= $\tan^{-1} \frac{36+20}{48-15}$
= $\tan^{-1} \frac{56}{33}$
Using equation(3)
= $\tan^{-1} \frac{56}{33}$
Hence proved

Question 6. Prove $\cos^{-1} \frac{12}{13}+\sin^{-1} \frac{3}{5}=\sin^{-1} \frac{56}{65}$

Solution:

Let $\sin^{-1} \frac{3}{5}=x$
Then, sin x = 3/5
$\cos x =\sqrt{1- (\frac{3}{5})^{2}}=\sqrt \frac{16}{25}$ = 4/5
$\therefore \tan x = \frac{3}{4} \implies x= \tan^{-1} \frac{3}{4}$
$\therefore \sin^{-1} \frac{3}{5}= \tan^{-1} \frac{3}{4}$ -(1)
Now, let $\cos^{-1} \frac{12}{13}=y$
Then, cos y = 12/13 and sin y = 5/13
$\therefore \tan y = \frac{5}{12} \implies y= \tan^{-1} \frac{5}{12}$
$\therefore \cos^{-1} \frac{12}{13}= \tan^{-1} \frac{5}{12}$ -(2)
Let $\sin^{-1} \frac{56}{65}=z$
Then, sin z = 56/65 and cos z = 33/65
$\therefore \tan z = \frac{56}{33} \implies z=\tan ^{-1} \frac{56}{33}$
$\therefore \sin^{-1} \frac{56}{65}= \tan^{-1} \frac{56}{33}$ -(3)
Now, we have:
L.H.S.= $\cos^{-1} \frac{12}{13}+ \sin^{-1} \frac{3}{5}$
From equation(1) and equation(2)
= $\tan^{-1} \frac{5}{12}+\tan^{-1} \frac{3}{4}$
= $\tan^{-1} \frac{{\frac{5}{12}+ \frac{3}{4}} }{1-{\frac{5}{12}\times \frac{3}{4}}}$ – $[\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$
= $\tan^{-1} \frac{20+36}{48-15}$
= $\tan^{-1} \frac{56}{33}$
From equation (3)
= $\sin^{-1} \frac{56}{65}$
Hence proved

Question 7. Prove $\tan^{-1} \frac{63}{16}= \sin^{-1} \frac{5}{13}+\cos^{-1} \frac{3}{5}$

Solution:

Let $\sin^{-1} \frac{5}{13}=x$
Then, sin x = 5/13 and cos x = 12/13.
$\tan^{-1} \frac{7+5}{35-1}+\tan^{-1} \frac{8+3}{24-1}$
$\therefore \tan x= \frac{5}{12} \to x= \tan^{-1} \frac{5}{12}$
$\therefore \sin^{-1} \frac{5}{13}= \tan^{-1} \frac{5}{12}$ -(1)
Let $\cos^{-1} \frac{3}{5}=y$
Then, cos y = 3/5 and sin y = 4/5
$\therefore \tan y= \frac{4}{3} \implies y= \tan^{-1}\frac{4}{3}$
$\therefore \cos ^{-1}\frac{3}{5}=\tan^{-1} \frac{4}{3}$ -(2)
From equation(1) and (2), we have
R.H.S. $=\sin^{-1} \frac{5}{13}+\cos^{-1} \frac{3}{5}$
= $\tan^{-1} \frac{5}{12}+\tan^{-1} \frac{4}{3}$
= $\tan^{-1} \frac{{\frac{5}{12}+ \frac{4}{3}} }{1-{\frac{5}{12}\times \frac{4}{3}}}$ – $[\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$
= $\tan^{-1} \frac{15+48}{36-20}$
= $\tan^{-1} \frac{63}{16}$
L.H.S = R.H.S
Hence proved

Question 8. Prove $\tan^{-1} \frac{1}{5}+\tan^{-1} \frac{1}{7}\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{8}=\frac{\pi}{4}$

Solution:

L.H.S. $=\tan^{-1} \frac{1}{5}+\tan^{-1} \frac{1}{7}\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{8}$
= $\tan^{-1} \frac{{\frac{1}{5}+ \frac{1}{7}} }{1-{\frac{1}{5}\times \frac{1}{7}}} +\tan^{-1} \frac{{\frac{1}{3}+ \frac{1}{8}} }{1-{\frac{1}{3}\times \frac{1}{8}}}$ – $[\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$
= $\tan^{-1} \frac{7+5}{35-1}+\tan^{-1} \frac{8+3}{24-1}$
= $\tan^{-1} \frac{12}{34}+\tan^{-1} \frac{11}{23}$
= $\tan^{-1} \frac{6}{17}+\tan^{-1} \frac{11}{23}$
= $\tan^{-1} \frac{{\frac{6}{17}+ \frac{11}{23}} }{1-{\frac{6}{17}\times \frac{11}{23}}}$
= $\tan^{-1} \frac{138 + 187}{391-66}$
= $\tan^{-1} \frac{325}{325}=\tan^{-1} 1$
= π/4
L.H.S = R.H.S
Hence proved

Question 9. Prove $\tan^{-1} \sqrt x= \frac{1}{2} \cos^{-1} (\frac{1-x}{1+x}),x\in[0,1]$

Solution:

Let x = tan²θ
Then, $\sqrt x=\tan \theta \implies\theta=\tan^{-1} \sqrt x.$
$\therefore \frac{1-x}{1+x}+\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\cos 2\theta$
Now, we have
R.H.S = $\frac{1}{2} \cos ^{-1}(\frac{1-x}{1+x})= \frac{1}{2} \cos ^{-1} (\cos 2 \theta)=\frac{1}{2} \times 2 \theta=\theta=\tan^{-1}\sqrt x$
L.H.S = R.H.S
Hence proved

Question 10. Prove $\cot^{-1} (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})=\frac{x}{2},x\in(0,\frac{\pi}{4})$

Solution:

Consider $(\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})$
By rationalizing
= $\frac{(\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x}))^{2}}{(\sqrt{ ({1+ \sin x})}-\sqrt({1- \sin x}))^{2}}$
= $\frac{( {1+ \sin x)} + {(1-\sin x)} + 2 \sqrt{(1+\sin x)(1-\sin x)}}{{ {1+ \sin x}}-{1+\sin x}}$
= $\frac{2(1+\sqrt{1-\sin^{2}})}{2\sin x}=\frac{1+\cos x}{\sin x}=\frac{2\cos ^{2}\frac{x}{2}}{2\sin \frac{x}{2}\cos\frac{x}{2}}$
= $\cot \frac{x}{2}$
L.H.S = $\cot^{-1} (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})={\cot}^{-1}({\cot( \frac x 2)})$ = x/2
L.H.S = R.H.S
Hence proved

Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 2

Last Updated : 05 Apr, 2021

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Applications of Derivatives

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Applications of Derivatives

Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 1

Question 1. Find the value of

Question 2. Find the value of

Question 3. Prove

Question 4. Prove

Question 5. Prove

Question 6. Prove

Question 7. Prove

Question 8. Prove

Question 9. Prove

Question 10. Prove

Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 2

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Question 1. Find the value of ${\cos }^{-1}(\cos \frac {13\pi} {6})$

Question 2. Find the value of $\tan^{-1}(\tan \frac {7\pi}{6})$

Question 3. Prove $2\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{24}{7}$

Question 4. Prove $\sin^{-1} \frac{8}{17}+\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{77}{36}$

Question 5. Prove $\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$

Question 6. Prove $\cos^{-1} \frac{12}{13}+\sin^{-1} \frac{3}{5}=\sin^{-1} \frac{56}{65}$

Question 7. Prove $\tan^{-1} \frac{63}{16}= \sin^{-1} \frac{5}{13}+\cos^{-1} \frac{3}{5}$

Question 8. Prove $\tan^{-1} \frac{1}{5}+\tan^{-1} \frac{1}{7}\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{8}=\frac{\pi}{4}$

Question 9. Prove $\tan^{-1} \sqrt x= \frac{1}{2} \cos^{-1} (\frac{1-x}{1+x}),x\in[0,1]$

Question 10. Prove $\cot^{-1} (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})=\frac{x}{2},x\in(0,\frac{\pi}{4})$