Class 12 NCERT Solutions Mathematics Part I Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise on Chapter 2 | Set 1 (original) (raw)
Last Updated : 23 Jul, 2025
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 = tan2_θ_
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