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

  • Last Updated : 05 Apr, 2021

Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 1 

Question 11. 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


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

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


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

Question 13. Solve 2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)


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


\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}

π/4 – θ = θ/2

θ = π/6

So, x = tan(π/6) = 1/√3

Question 15. 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})}}


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 16. 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 


\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


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

Question 17. 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  


\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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