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

Content of this article has been merged with Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise as per the revised syllabus of NCERT.

Question 11. Prove [Tex]\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[/Tex]

Solution:

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

Then, we have :

LHS = [Tex]\tan^{-1}(\frac{\sqrt{(1+ x) }-\sqrt{(1-x)}}{\sqrt{(1+ x )}+\sqrt{(1-x)}})[/Tex]

= [Tex]\tan^{-1}(\frac{\sqrt{(1+ \cos2 \theta) }-\sqrt{(1-\cos2 \theta)}}{\sqrt{(1+ \cos2 \theta) }+\sqrt{(1-\cos2 \theta)}})[/Tex]

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

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

= [Tex]\tan^{-1}(\frac{\cos \theta – \sin \theta}{\cos \theta + \sin \theta})=\tan^{-1}(\frac{1 – \tan \theta}{1 + \tan \theta})[/Tex]

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

[Tex]=\frac{\pi}{4} -\theta = \frac{\pi}{4}-\frac{1}{2} \cos^{-1} x      [/Tex]  

L.H.S = R.H.S

Hence Proved

Question 12. Prove [Tex]\frac{9\pi}{8}- \frac{9}{4} \sin^{-1} \frac{1}{3}=\frac{9}{4} \sin^{-1} \frac{2\sqrt2}{3}[/Tex]

Solution:

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

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

Using [Tex]\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}[/Tex]

= [Tex]\frac {9}{4}(\cos^{-1} \frac{1}{3})      [/Tex]      -(1)

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

[Tex]\therefore x=\sin^{-1} \frac{2\sqrt2}{3} [/Tex] 

Using equation(1), we get,

= [Tex]\frac{9}{4}\sin^{-1} \frac{2\sqrt2}{3}[/Tex]

L.H.S = R.H.S

Hence Proved

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

Solution:

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

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

= [Tex]\frac{2\cos x} {1-cos^{2}x}=2\cosec x[/Tex]

= [Tex]\frac{ 2 \cos x}{\sin^{2}}= \frac{2}{\sin x}[/Tex]

= cos x/sin x

= cot x =1

Therefore, x = π/4​

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

Solution:

[Tex]\tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x[/Tex]

Let x = tanθ

[Tex]\tan^{-1}(\frac{1-tan\theta}{1+tan\theta})=\frac{1}{2} \tan^{-1} tan\theta     [/Tex]        

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

[Tex]\tan^{-1}tan(\frac{\pi}{4}-\theta)=\frac{\theta }{2}    [/Tex]

π/4 – θ = θ/2

θ = π/6

So, x = tan(π/6) = 1/√3

Question 15. Solve [Tex]\sin(\tan^{-1}x),|x|<1 [/Tex] is equal to

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

Solution:

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

Let [Tex]\tan^{-1} x=y [/Tex] Then,

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

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

So, the correct answer is D.

Question 16. Solve [Tex]\sin^{-1}(1- x)-2 \sin^{-1}x=\frac{\pi}{2} [/Tex], then x is equal to

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

Solution:

[Tex]\sin^{-1}(1- x)-2 \sin^{-1}x=\frac{\pi}{2}[/Tex]

[Tex]\implies -2 \sin^{-1}x=\frac{\pi}{2} – \sin^{-1}(1-x)[/Tex]

[Tex]\implies -2 \sin^{-1}x=\cos^{-1}(1-x)     [/Tex]        -(1)

Let [Tex]\sin^{-1} x =\theta \to \sin \theta=x [/Tex]

[Tex]\cos \theta= \sqrt{1-x^{2}}[/Tex]

[Tex]\therefore \theta= \cos^{-1}(\sqrt{1-x^{2}})[/Tex]

[Tex]\therefore \sin^{-1} x=cos^{-1}(\sqrt{1-x^{2}})[/Tex]

Therefore, from equation(1), we have

[Tex]-2cos^{-1}(\sqrt{1-x^{2}})=\cos^{-1}(1-x)[/Tex]

Put x = siny then, we have:

[Tex]-2\cos^{-1}(\sqrt{1-\sin^{2} y})=\cos^{-1}(1-\sin y)[/Tex]

[Tex]-2 \cos^{-1}(\cos y)=\cos^{-1}(1-\sin y)[/Tex]

[Tex]-2 y=\cos^{-1}(1-\sin y)[/Tex]

[Tex]1- \sin y=\cos (-2y)=\cos 2y[/Tex]

[Tex]1- \sin y= 1- 2 \sin^{2} y[/Tex]

[Tex]2\sin^{2} y- \sin y=0[/Tex]

[Tex]\sin y(2 \sin y-1)=0[/Tex]

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. = [Tex]\sin^{-1}(1-\frac{1}{2}) -2\sin^{-1} \frac{1}{2}[/Tex]

= [Tex]\sin^{-1} (\frac{1}{2})-2\sin^{-1} \frac{1}{2}[/Tex]

= [Tex]-\sin^{-1} \frac{1}{2}[/Tex]

= [Tex]- \frac{\pi}{6} \ne\frac{\pi}{2}\ne \space R.H.S.[/Tex]

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

Thus, x = 0 

Hence, the correct answer is C

Question 17. Solve [Tex]\tan ^{-1}(\frac{x}{y})-\tan ^{-1}(\frac{x-y}{x+y}) [/Tex] is equal to

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

Solution

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

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

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

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

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

Hence, the correct answer is C