Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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

  • Last Updated : 05 Apr, 2021

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

Solution: 

We know that \cos^{-1} (\cos x)=x  

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

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




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!