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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  

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

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