Introduction
Preparing for CBSE Class 12 board exams can feel overwhelming, especially when it comes to mastering Inverse Trigonometry in Maths. Miscellaneous Exercise from NCERT Class 12 Maths Chapter 2 – Inverse Trigonometry is one of the first steps toward understanding this important topic.
Have been looking for solution to Miscellaneous exercise from chapter 2 of Class 12th Maths NCERT ? This post gives you step-by-step NCERT solutions for every question in the Miscellaneous Exercise, explained clearly and formatted just like your textbook. Whether you’re a student looking to improve your concepts or a teacher searching for reliable solutions to share with your class, these answers will make integration simpler and more approachable.
By the end of this page, you’ll build confidence for solving more complex inverse trigonometric problems in later exercises.
Let’s get started and make Class 12 Maths Miscellaneous Exercise from Chapter 2 easy to learn — and easy to score in exams!
Question 1: Evaluate: \cos^{-1}(\cos(\frac{13\pi}{6}))
Solution:
\cos(\frac{13\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}
So,
\cos^{-1}(\cos(\frac{13\pi}{6})) = \cos^{-1}(\frac{\sqrt{3}}{2})
\cos^{-1}(\frac{\sqrt{3}}{2}) = \cos^{-1}(\cos\frac{\pi}{6})
\cos^{-1}(\cos\frac{\pi}{6}) = \frac{\pi}{6}
Final Answer: \boxed{ \cos^{-1}(\cos\frac{13\pi}{6}) = \frac{\pi}{6} }
Question 2: Evaluate: \tan^{-1}(\tan(\frac{7\pi}{6}))
Solution:
\tan(\frac{7\pi}{6}) = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}
So,
\tan^{-1}(\tan(\frac{7\pi}{6})) = \tan^{-1}(\frac{1}{\sqrt{3}}))
\tan^{-1}(\frac{1}{\sqrt{3}}) = \tan^{-1}(\tan(\frac{\pi}{6}))
\tan^{-1}(\tan(\frac{\pi}{6})) = \frac{\pi}{6}
Final Answer: \boxed{ \tan^{-1}(\tan(\frac{7\pi}{6})) = \frac{\pi}{6} }
Question 3: Prove that: 2\sin^{-1}(\frac{3}{5}) = \tan^{-1}(\frac{24}{7})
Solution:
Let \theta = \sin^{-1}(\frac{3}{5}),
then, \sin \theta = \frac{3}{5}, \tan \theta = \frac{3}{4},
Using the identity, \tan (2 \theta) = \frac {2 \tan \theta}{1- \tan^2 \theta}
So,
\tan 2\theta = \frac{2 \cdot 3/4}{1 - (3/4)^2}
\tan 2\theta = \frac{24}{7}
2\theta = \tan^{-1}(\frac{24}{7})
\boxed { 2\sin^{-1}(\frac{3}{5}) = \tan^{-1}(\frac{24}{7}) } Hence Proved.
Question 4: Prove that: \sin^{-1}(\frac{8}{17}) + \sin^{-1}(\frac{3}{5}) = \tan^{-1}(\frac{77}{36})
Solution:
Let \alpha = \sin^{-1}(\frac{8}{17}), \beta = \sin^{-1}(\frac{3}{5})
\sin \alpha = \frac{8}{17}, \tan \alpha = \frac{8}{15}
\sin \beta = \frac{3}{5}, \tan \beta = \frac{3}{4}
Use angle addition for tan: \tan(\alpha + \beta) = \frac{\tan \alpha +\tan \beta}{1 - \tan \alpha \cdot \tan \beta}
\tan(\alpha + \beta) = \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \cdot \frac{3}{4}}
\tan(\alpha + \beta) = \frac{77}{36}
\alpha + \beta = \tan^{-1}(\frac{77}{36})
\boxed{ \sin^{-1}(\frac{8}{17}) + \sin^{-1}(\frac{3}{5}) = \tan^{-1}(\frac{77}{36}) } Hence Proved.
Question 5: Prove that: \cos^{-1}(\frac{4}{5}) + \cos^{-1}(\frac{12}{13}) = \cos^{-1}(\frac{33}{65})
Solution:
Let, A = \cos^{-1}(\frac{4}{5}) and B = \cos^{-1}(\frac{12}{13})
\cos A = \frac{4}{5} , \quad \sin A = \frac{3}{5}
\cos B = \frac{12}{13} , \quad \sin B = \frac{5}{13}
Using formula \cos (A+B) = \cos A \cos B - \sin A \sin B
\cos (A+B) = \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13}
\cos (A+B) = \frac{48}{65} - \frac{15}{65}
\cos (A+B) = \frac{33}{65}
A+B = \cos^{-1}(\frac{33}{65})
\boxed { \cos^{-1}(\frac{4}{5}) + \cos^{-1}(\frac{12}{13}) = \cos^{-1}(\frac{33}{65}) } Hence Proved
Question 6: Prove that: \cos^{-1}(\frac{12}{13}) + \sin^{-1}(\frac{3}{5}) = \sin^{-1}(\frac{56}{65})
Solution:
Let, A = \cos^{-1}(\frac{12}{13}) and B = \sin^{-1}(\frac{3}{5})
\cos A = \frac{12}{13} , \quad \sin A = \frac{5}{13}
\sin B = \frac{3}{5} , \quad \cos B = \frac{4}{5}
Using formula \sin (A+B) = \sin A \cos B + \cos A \sin B
\sin (A+B) = \frac{5}{13} \cdot \frac{4}{5} + \frac{12}{13} \cdot \frac{3}{5}
\sin (A+B) = \frac{20}{65} + \frac{36}{65}
\sin (A+B) = \frac{56}{65}
A+B = \sin^{-1} (\frac{56}{65})
\boxed{ \cos^{-1}(\frac{12}{13}) + \sin^{-1}(\frac{3}{5}) = \sin^{-1}(\frac{56}{65}) } Hence Proved
Question 7: Prove that: \tan^{-1}(\frac{63}{16}) = \sin^{-1}(\frac{5}{13}) + \cos^{-1}(\frac{3}{5})
Solution:
Let, A = \sin^{-1}(\frac{5}{13}) and B = \cos^{-1}(\frac{3}{5})
\sin A = \frac{5}{13} , \quad \tan A = \frac{5}{12}
\cos B = \frac{3}{5} , \quad \tan B = \frac{4}{3}
Using, \tan (A+B) = \frac{\tan A +\tan B}{1 - \tan A \cdot \tan B}
\tan (A+B) = \frac{ \frac{5}{12} + \frac{4}{3} } { 1 - \frac{5}{12} \cdot \frac{4}{3}}
\tan (A+B) = \frac{ \frac{63}{36}} { 1 - \frac{20}{36}}
\tan (A+B) = \frac{ \frac{63}{36}} {\frac{16}{36}}
\tan (A+B) = \frac{63}{16}
A+B = \tan^{-1}(\frac{63}{16})
\sin^{-1}(\frac{5}{13}) + \cos^{-1}(\frac{3}{5}) = \tan^{-1}(\frac{63}{16})
\boxed{ \tan^{-1}(\frac{63}{16}) = \sin^{-1}(\frac{5}{13}) + \cos^{-1}(\frac{3}{5}) } Hence Proved
Question 8: Prove that: \tan^{-1}(\sqrt{x}) = \frac{1}{2} \cos^{-1} \left( \frac{1 - x}{1 + x} \right), \quad \text{for } x \in [0, 1]
Solution:
Let \theta = \tan^{-1}(\sqrt{x})
Then, \tan \theta = \sqrt{x} \Rightarrow x = \tan^2 \theta
Now use the identity: \cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}
Substituting \tan^2 \theta = x,
we get: \cos(2\theta) = \frac{1 - x}{1 + x}
Taking \cos^{-1} on both sides:
2\theta = \cos^{-1} \left( \frac{1 - x}{1 + x} \right)
So, \theta = \frac{1}{2} \cos^{-1} \left( \frac{1 - x}{1 + x} \right)
But \theta = \tan^{-1}(\sqrt{x})
\boxed{ \tan^{-1}(\sqrt{x}) = \frac{1}{2} \cos^{-1} \left( \frac{1 - x}{1 + x} \right)} . Hence Proved
Question 9: Prove that: \cot^{-1} \left( \frac{ \sqrt{1 + \sin x} + \sqrt{1 - \sin x} }{ \sqrt{1 + \sin x} - \sqrt{1 - \sin x} } \right) = \frac{x}{2} , \quad \text{for } x \in \left( 0, \frac{\pi}{4} \right)
Solution:
LHS = \cot^{-1} \left( \frac{ \sqrt{1 + \sin x} + \sqrt{1 - \sin x} }{ \sqrt{1 + \sin x} - \sqrt{1 - \sin x} } \right)
LHS = \cot^{-1} \left( \frac{ (\sqrt{1 + \sin x} + \sqrt{1 - \sin x}) ((\sqrt{1 + \sin x} + \sqrt{1 - \sin x})) }{ (\sqrt{1 + \sin x} - \sqrt{1 - \sin x})((\sqrt{1 + \sin x} + \sqrt{1 - \sin x})) } \right)
LHS = \cot^{-1} \left( \frac{ (\sqrt{1 + \sin x} + \sqrt{1 - \sin x})^2}{ (\sqrt{1 + \sin x})^2 - (\sqrt{1 - \sin x})^2} \right)
LHS = \cot^{-1} \left(\frac{ 1 + \sin x + 1 - \sin x + 2\cdot\sqrt{(1 + \sin x)(1 - \sin x)})}{ (1 + \sin x) - (1 - \sin x)} \right)
LHS = \cot^{-1} \left(\frac{ 2 + 2\cdot\sqrt{(1 - \sin^2 x)}}{ 2\sin x} \right)
LHS = \cot^{-1} \left(\frac{ 2 + 2\cdot\cos x}{ 2\sin x} \right)
LHS = \cot^{-1} \left(\frac{ 2 (1 + \cdot\cos x)}{ 2\sin x} \right)
LHS = \cot^{-1} \left(\frac{ (1 + \cdot\cos x)}{\sin x} \right)
LHS = \cot^{-1} \left(\frac{ 2 \cos^2{\frac{x}{2}}}{ 2\sin \frac{x}{2} \cdot \cos \frac{x}{2} } \right)
LHS = \cot^{-1} \left(\frac{ \cos{\frac{x}{2}}}{ 2\sin \frac{x}{2}} \right)
LHS = \cot^{-1} \left(\cot{\frac{x}{2}} \right)
LHS = \frac{x}{2} = RHS
\boxed{ \cot^{-1} \left( \frac{ \sqrt{1 + \sin x} + \sqrt{1 - \sin x} }{ \sqrt{1 + \sin x} - \sqrt{1 - \sin x} } \right) = \frac{x}{2} } Hence Proved
Question 10: Prove that: \tan^{-1} \left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1}(x), \quad \text{where } -\frac{1}{\sqrt{2}} \le x \le 1
Solution:
Let x = \cos \theta
Plugging x in \left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right)
We get, \left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right) = \left( \frac{ \sqrt{1 + \cos \theta} - \sqrt{1 - \cos \theta} }{ \sqrt{1 + \cos \theta} + \sqrt{1 - \cos \theta} } \right)
Using 1 + \cos \theta = 2 \cos^2 {\frac{\theta}{2}} and 1 - \cos \theta = 2 \sin^2 {\frac{\theta}{2}}
\left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right) = \left( \frac{ \sqrt{2 \cos^2 {\frac{\theta}{2}}} - \sqrt{2 \sin^2 {\frac{\theta}{2}}} }{ \sqrt{2 \cos^2 {\frac{\theta}{2}}} + \sqrt{2 \sin^2 {\frac{\theta}{2}}} } \right)
\left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right) = \left( \frac{ \cos {\frac{\theta}{2}} - \sin {\frac{\theta}{2}} }{ \cos {\frac{\theta}{2}} + \sin {\frac{\theta}{2}} } \right)
Dividing numerator and denominator by \cos {\frac{\theta}{2}}
\left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right) = \left( \frac{ 1 - \tan {\frac{\theta}{2}} }{ 1 + \tan {\frac{\theta}{2}} } \right)
\left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right) = \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right)
\tan^{-1} \left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right) = \left( \frac{\pi}{4} - \frac{\theta}{2} \right)
Substituting back \theta = \cos^{-1} x
\tan^{-1} \left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right) = \frac{\pi}{4} - \frac{1}{2} \cdot \cos^{-1} x
\boxed{ \tan^{-1} \left( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \right) = \frac{\pi}{4} - \frac{1}{2} \cdot \cos^{-1} x } . Hence Proved
Question 11: Solve: 2\tan^{-1}(\cos x) = \tan^{-1}(2\cosec x)
Solution:
Let \theta = \tan^{-1}(\cos x)
Therefore, equation becomes 2 \theta = \tan^{-1}(2\cosec x)
\tan {2 \theta} = 2\cosec x
\frac {2 \tan {\theta}}{1 - \tan^2 {\theta}} = 2\cosec x
Substituting \tan \theta = \cos x from above,
\frac {2 \cos x}{1 - \cos^2 {x}} = 2\cosec x
\frac {2 \cos x}{\sin^2 {x}} = 2\cosec x
2 \cos x = 2 \cosec x \cdot \sin ^2 x
\cos x = \sin x
\cot x = 1
Final Answer: \boxed{ x = n\pi + \frac{\pi}{4} , \quad n \in Z }
Question 12: Solve: \tan^{-1}\left(\frac{1 - x}{1 + x} \right) = \frac{1}{2}\tan^{-1}(x), where x > 0
Solution:
Let, x = \tan \theta
\tan^{-1}\left(\frac{1 - \tan \theta}{1 + \tan \theta} \right) = \frac{1}{2}\tan^{-1}(x)
\tan^{-1}\left(\frac{\tan \frac{\pi}{4} - \tan \theta}{1 + \tan \frac{\pi}{4} \cdot \tan \theta} \right) = \frac{1}{2}\tan^{-1}(x)
\tan^{-1}\left(\tan (\frac{\pi}{4} - \theta) \right) = \frac{1}{2}\tan^{-1}(x)
\frac{\pi}{4} - \theta = \frac{1}{2}\tan^{-1}(x)
\frac{\pi}{2} - 2\theta = \tan^{-1}(x)
\tan (\frac{\pi}{2} - 2\theta) = x
\cot 2\theta = x
Using tan (2\theta) = \frac{2 \tan \theta}{1 - tan^2 \theta} \implies cot (2\theta) = \frac{1 - tan^2 \theta}{2 \tan \theta}
\frac{1 - \tan^2 \theta}{2 \tan \theta} = x
Substituting back \tan \theta = x
\frac{1 - x^2}{2x} = x
1 - x^2 = 2x^2
1 = 3 x^2
x^2 = \frac{1}{3}
x = \pm\frac{1}{\sqrt{3}}
Given x > 0
Final Answer: \boxed{x = \frac{1}{\sqrt{3}}}
Choose the correct option in the following questions
Question 13: Evaluate: \sin(\tan^{-1}(x)), for |x| < 1
(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 \theta = \tan^{-1}(x)
\tan \theta = x = \frac{x}{1}
Then, hypotenuse = \sqrt{1 + x^2}, perpendicular = x
\sin \theta = \frac{x}{\sqrt{1 + x^2}}
\theta = \sin^{-1} (\frac{x}{\sqrt{1 + x^2}})
So, \sin(\tan^{-1}(x)) = \sin \theta
\sin(\tan^{-1}(x)) = \sin \big( \sin^{-1} (\frac{x}{\sqrt{1 + x^2}}) \big)
\boxed{ \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}} }
Correct Option: (D)
Question 14: If \sin^{-1}(1 - x) - 2\sin^{-1}(x) = \frac{\pi}{2}, then find the value of x
(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}
\sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1}(x)
(1 - x) = \sin \big( \frac{\pi}{2} + 2\sin^{-1}(x) \big)
(1 - x) = \cos \big( 2\sin^{-1}x \big)
(1 - x) = 1 - 2 \sin^2 \big(\sin^{-1}x \big)
1 - x = 1 - 2 \cdot x^2
2x^2 - x = 0
x(2x - 1) = 0
x = 0, \quad x = \frac{1}{2}
Try x = 0:
latex]\sin^{-1}(1 – 0) – 2\sin^{-1}(0) = \sin^{-1}(1) = \frac{\pi}{2}[/latex] ✅
Try x = \frac{1}{2}:
\sin^{-1}(1 - \frac{1}{2}) - 2\cdot \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} ✖
Only x = 0 satisfies the equation.
Correct Option: (C)