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

Last Updated : 03 Apr, 2024
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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})  = Ï€/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})  = Ï€/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 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 9. 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

Question 10. Prove \tan^{-1}(\frac{\sqrt{(1+ x )}-\sqrt{(1-x)}}{\sqrt{(1+ x) }+\sqrt{(1-x)}})=\frac{\pi}{4}-\frac{1}{2} \cos^{-1}x,-\frac{1}{\sqrt2}\le x \le1

Solution:

Put x=\cos2\theta  so that, \theta= \frac{1}{2} \cos^{-1} x.

Then, we have :

LHS = \tan^{-1}(\frac{\sqrt{(1+ x) }-\sqrt{(1-x)}}{\sqrt{(1+ x )}+\sqrt{(1-x)}})

= \tan^{-1}(\frac{\sqrt{(1+ \cos2 \theta) }-\sqrt{(1-\cos2 \theta)}}{\sqrt{(1+ \cos2 \theta) }+\sqrt{(1-\cos2 \theta)}})

= \tan^{-1}(\frac{\sqrt{(2 \cos^{2} \theta) }-\sqrt{(2 \sin^{2}\theta)}}{\sqrt{(2 \cos^{2}\theta) }+\sqrt{(2 \sin^{2} \theta)}})

= \tan^{-1}(\frac{\sqrt{2 }\cos \theta -{\sqrt{2 }\cos\theta}}{{\sqrt{2 }\cos\theta }+{\sqrt{2 }\cos \theta}})

= \tan^{-1}(\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta})=\tan^{-1}(\frac{1 - \tan \theta}{1 + \tan \theta})

\tan^{-1} 1- \tan^{-1}(\tan \theta)       – [\tan^{-1}(\frac{x-y}{1+xy})]=\tan^{-1}x+\tan^{-1}y

=\frac{\pi}{4} -\theta = \frac{\pi}{4}-\frac{1}{2} \cos^{-1} x   

L.H.S = R.H.S

Hence Proved

Question 11. Solve 2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)

Solution:

2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)

= \tan^{-1}(\frac{2 \cos x}{1- \cos^{2}x})=\tan^{-1} (2\cosec x)                       –[2 \tan^{-1} x=\tan^{-1} \frac{2x}{1-x^{2}}]

= \frac{2\cos x} {1-cos^{2}x}=2\cosec x

= \frac{ 2 \cos x}{\sin^{2}}= \frac{2}{\sin x}

= cos x/sin x

= cot x =1

Therefore, x = π/4​

Question 12. Solve \tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x,(x>0)

Solution:

\tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x

Let x = tanθ

\tan^{-1}(\frac{1-tan\theta}{1+tan\theta})=\frac{1}{2} \tan^{-1} tan\theta         

\tan^{-1}(\frac{tan\frac{\pi}{4}-tan\theta}{tan\frac{\pi}{4}+tan\theta})=\frac{1}{2} \theta

\tan^{-1}tan(\frac{\pi}{4}-\theta)=\frac{\theta }{2}

Ï€/4 – θ = θ/2

θ = π/6

So, x = tan(π/6) = 1/√3

Question 13. Solve \sin(\tan^{-1}x),|x|<1  is equal to

(A) \frac{x}{\sqrt{(1-x^{2})}}     (B) \frac{1}{\sqrt{(1-x^{2})}}  (C) \frac{1}{\sqrt{(1+x^{2})}} (D) \frac{x}{\sqrt{(1+x^{2})}}

Solution:

Let tan y = x, \sin y = \frac{x}{\sqrt{(1+x^{2})}}

Let \tan^{-1} x=y  Then,

\therefore y=\sin^{-1}(\frac{x}{\sqrt{(1+x^{2})}}) \implies \tan^{-1}x=\sin^{-1}\frac{x}{\sqrt{(1+x^{2})}}

\therefore \sin(\tan^{-1}x)=sin(sin^{-1}\frac{x}{\sqrt{(1+x^{2})}})=\frac{x}{\sqrt{(1+x^{2})}}

So, the correct answer is D.

Question 14. Solve \sin^{-1}(1- x)-2 \sin^{-1}x=\frac{\pi}{2} , then x is equal to

(A) 0, 1/2      (B) 1, 1/2      (C) 0     (D) 1/2 

Solution:

\sin^{-1}(1- x)-2 \sin^{-1}x=\frac{\pi}{2}

\implies -2 \sin^{-1}x=\frac{\pi}{2} - \sin^{-1}(1-x)

\implies -2 \sin^{-1}x=\cos^{-1}(1-x)         -(1)

Let \sin^{-1} x =\theta \to \sin \theta=x

\cos \theta= \sqrt{1-x^{2}}

\therefore \theta= \cos^{-1}(\sqrt{1-x^{2}})

\therefore \sin^{-1} x=cos^{-1}(\sqrt{1-x^{2}})

Therefore, from equation(1), we have

-2cos^{-1}(\sqrt{1-x^{2}})=\cos^{-1}(1-x)

Put x = siny then, we have:

-2\cos^{-1}(\sqrt{1-\sin^{2} y})=\cos^{-1}(1-\sin y)

-2 \cos^{-1}(\cos y)=\cos^{-1}(1-\sin y)

-2 y=\cos^{-1}(1-\sin y)

1- \sin y=\cos (-2y)=\cos 2y

1- \sin y= 1- 2 \sin^{2} y

2\sin^{2} y- \sin y=0

\sin y(2 \sin y-1)=0

sin y = 0 or 1/2

x = 0 or x = 1/2

But, when x = 1/2 it can be observed that:

L.H.S. = \sin^{-1}(1-\frac{1}{2}) -2\sin^{-1} \frac{1}{2}

= \sin^{-1} (\frac{1}{2})-2\sin^{-1} \frac{1}{2}

= -\sin^{-1} \frac{1}{2}

= - \frac{\pi}{6} \ne\frac{\pi}{2}\ne \space R.H.S.

x = 1/2 is not the solution of given equation.

Thus, x = 0 

Hence, the correct answer is C

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

= Ï€/4 

L.H.S = R.H.S

Hence proved

Prove \frac{9\pi}{8}- \frac{9}{4} \sin^{-1} \frac{1}{3}=\frac{9}{4} \sin^{-1} \frac{2\sqrt2}{3}

Solution:

L.H.S. = \frac{9\pi}{8}- \frac{9}{4} \sin^{-1} \frac{1}{3}

= \frac {9}{4}(\frac{\pi}{2}- \sin^{-1} \frac{1}{3})

Using \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}

= \frac {9}{4}(\cos^{-1} \frac{1}{3})       -(1)

Now, let \cos^{-1}\frac{1}{3}=x  Then, \cos x =\frac{1}{3} \implies \sin x=\sqrt{1-(\frac{1}{3})^{2}}=\frac{2\sqrt{2}}{3}

\therefore x=\sin^{-1} \frac{2\sqrt2}{3}  

Using equation(1), we get,

= \frac{9}{4}\sin^{-1} \frac{2\sqrt2}{3}

L.H.S = R.H.S

Hence Proved

Solve \tan ^{-1}(\frac{x}{y})-\tan ^{-1}(\frac{x-y}{x+y})  is equal to

(A) π​/2      (B) π​/3     (C) π​/4      (D) -3π​/4  

Solution

\tan^{-1} (\frac{x}{y})-\tan^{-1} \frac{x-y}{x+y}

\tan^{-1}[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y} \times\frac{x-y}{x+y}}]                 –[\tan^{-1}x+\tan^{-1}y=[\tan^{-1}(\frac{x-y}{1+xy})]]

\tan^{-1}[\frac {\frac {x(x+y)-y(x-y)} {y(x+y)} } {\frac {y(x+y)+x(x-y)} {y(x+y)}}]

{\tan}^{-1}[{\frac {x^2+xy-xy+y^2} {xy+y^2+x^2-xy}}]

{\tan}^{-1}[\frac {x^2+y^2} {x^2+y^2}]=tan^{-1}1=\frac {\pi} {4}

Hence, the correct answer is C



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