Class 12 NCERT Solutions Mathematics Part I Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise on Chapter 2 | Set 2 (original) (raw)
Last Updated : 23 Jul, 2025
Chapter 2 of Class 12 NCERT Mathematics Part I delves deeper into Inverse Trigonometric Functions, building upon the foundational concepts introduced earlier. This set of miscellaneous exercises challenges students to apply their understanding of inverse trigonometric functions in more complex scenarios. It emphasizes problem-solving skills, requiring students to manipulate and combine various inverse trigonometric identities, properties, and formulas. These exercises are designed to enhance students' analytical thinking and prepare them for advanced mathematical concepts in higher studies.
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 \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:
Putx=\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 12. 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
Question 13. 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 14. 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 = tan__θ_
\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})}}
**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 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
**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
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) -3__π_/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
Summary
This set of miscellaneous exercises on Inverse Trigonometric Functions reinforces students' understanding of these complex mathematical concepts. It covers a wide range of problem types, including proving identities, solving equations involving multiple inverse trigonometric functions, and evaluating compound expressions. The exercises require students to apply various techniques such as algebraic manipulation, trigonometric identities, and properties of inverse functions. By working through these problems, students develop a deeper intuition for the behavior of inverse trigonometric functions and their interrelationships, preparing them for more advanced mathematical analysis in fields like calculus and physics.