Introduction to Chapter 2: Inverse Trigonometric Functions

NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution lays the foundation of Mathematics. This chapter covers basic introduction of inverse functions of Trigonometry such as Sine, Cosine and tangent in the Trigonometry along with their properties and domain and range. Studying this is quite important for understanding these concepts, both from the point of view of board exams and competitive exams like JEE and other entrance exams, as many questions come from this chapter.

NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solutions

Q. Find the value of the following:

Q.1. NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution

\begin{gather*} cos^{-1}\left( cos\frac{13\pi }{6}\right) =cos^{-1}\left[ cos\left( 2\pi +\frac{\pi }{6}\right)\right]\\ =cos^{-1}\left( cos\frac{\pi }{6}\right) =\frac{\pi }{6}\\ Ans.\ \frac{\pi }{6} \end{gather*}

Q.2. NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution

\begin{array}{l} tan^{-1}\left( tan\frac{7\pi }{6}\right) =tan^{-1}\left[ tan\left( \pi +\frac{\pi }{6}\right)\right]\\ =tan^{-1}\left( tan\frac{\pi }{6}\right) =\frac{\pi }{6}\\ Ans.\ \frac{\pi }{6} \end{array}

Q.3. NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution

Q. Prove that:

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Q.3.\ Prove\ that:\ 2sin^{-1}\frac{3}{5} =tan^{-1}\frac{24}{7}\\
LHS=2sin^{-1}\frac{3}{5}\\
Let\ sin^{-1}\frac{3}{5} =x\ then\ sinx=\frac{3}{5}\\
We\ have,\ tanx=\frac{3}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\Longrightarrow x=\ tan^{-1}\frac{3}{4}\\
So,\ LHS=\ 2\ tan^{-1}\frac{3}{4}\\
=\ tan^{-1}\frac{2\left(\frac{3}{4}\right)}{1-\left(\frac{3}{4}\right)^{2}}\\
=tan^{-1}\frac{\frac{6}{4}}{1-\frac{9}{16}} =tan^{-1}\frac{\frac{6}{4}}{\frac{16-9}{16}}\\
=tan^{-1}\frac{\frac{6}{4}}{\frac{7}{16}} =tan^{-1}\left(\frac{6}{4} \times \frac{16}{7}\right)\\
=tan^{-1}\frac{24}{7}\\
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Q.4. NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution

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Q.4.\ Prove\ that:\\
sin^{-1}\frac{8}{17} +sin^{-1}\frac{3}{5} =tan^{-1}\frac{77}{36}\\
Solution:\ LHS=sin^{-1}\frac{8}{17} +sin^{-1}\frac{3}{5}\\
converting\ sin^{-1}\frac{8}{17} \ and\ sin^{-1}\frac{3}{5} \ into\ tan^{-1}( x)\\
LHS=\ tan^{-1}\frac{8}{15} +tan^{-1}\frac{3}{4}\\
Using\ Identity:\ tan^{-1} x+tan^{-1} y=tan^{-1}\left(\frac{x+y}{1-xy}\right)\\
=tan^{-1}\left(\frac{\frac{8}{15} +\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}\right)\\
=tan^{-1}\left(\frac{\frac{32+45}{60}}{\frac{60-24}{60}}\right)\\
=tan^{-1}\frac{77}{36} \ Ans.
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Q.5. NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution

Q.5. Prove that: $\displaystyle \begin{array}{{>{\displaystyle}l}} cos^{-1}\frac{4}{5} +cos^{-1}\frac{12}{13} =cos^{-1}\frac{33}{65}\\ Solution:\ LHS=\ cos^{-1}\frac{4}{5} +cos^{-1}\frac{12}{13}\\ Take\ cos^{-1}\frac{4}{5} =\alpha \ then,\ cos\alpha =\frac{4}{5}\\ so,\ tan\alpha =\frac{3}{4} \Longrightarrow \alpha =tan^{-1}\frac{3}{4}\\ Now,\ take\ cos^{-1}\frac{12}{13} =\beta ,\ then\ cos\beta =\frac{12}{13}\\ So,\ tan\beta =\frac{5}{12} \Longrightarrow \beta =tan^{-1}\frac{5}{12}\\ Substituting\ these\ values,\ we\ get\\ LHS=\ tan^{-1}\frac{3}{4} +tan^{-1}\frac{5}{12}\\ Using\ identity:\ tan^{-1} x+tan^{-1} y=tan^{-1}\left(\frac{x+y}{1-xy}\right)\\ =tan^{-1}\left(\frac{\frac{3}{4} +\frac{5}{12}}{1-\frac{3}{4} \times \frac{5}{12}}\right)\\ =tan^{-1}\left(\frac{\frac{36+20}{48}}{\frac{48-15}{48}}\right)\\ =tan^{-1}\frac{56}{33}\\ Converting\ into\ cos^{-1}( x)\\ tan^{-1}\frac{56}{33} =cos^{-1}\frac{33}{65} \ Ans. \end{array}$

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Q.6. NCERT Solutions for Class 12 Maths chapter 2 Miscellaneous Exercise

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Q.6.\ tan^{-1}\frac{x}{\sqrt{a^{2} -x^{2}}} ,\ |x|< a\\
Solution:\\
Put\ x=asin\theta \Longrightarrow sin\theta =\frac{x}{a}\\
\Longrightarrow \theta =sin^{-1}\frac{x}{a}\\
tan^{-1}\frac{asin\theta }{\sqrt{a^{2} -a^{2} sin^{2} \theta }}\\
=tan^{-1}\frac{asin\theta }{\sqrt{a^{2}\left( 1-sin^{2} \theta \right)}}\\
=tan^{-1}\frac{asin\theta }{acos\theta }\\
=tan^{-1}( tan\theta ) =\theta \\
=sin^{-1}\frac{x}{a}
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Q.7. NCERT Solutions for Class 12 Maths chapter 2 Miscellaneous Exercise

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Q.7.\ tan^{-1}\frac{63}{16} =sin^{-1}\frac{5}{13} +cos^{-1}\frac{3}{5}\\
Solution:\\
RHS=sin^{-1}\frac{5}{13} +cos^{-1}\frac{3}{5}\\
=tan^{-1}\frac{5}{12} +tan^{-1}\frac{4}{3}\\
=tan^{-1}\left(\frac{\frac{5}{12} +\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}\right)\\
=tan^{-1}\left(\frac{\frac{15+48}{36}}{\frac{36-20}{36}}\right)\\
=tan^{-1}\frac{\frac{63}{36}}{\frac{16}{36}}\\
=tan^{-1}\frac{63}{36} \ Ans.
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Q.8.NCERT Solutions for Class 12 Maths chapter 2 Miscellaneous Exercise

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Q.8.\ Prove\ that:\ tan^{-1}\sqrt{x} =\frac{1}{2} cos^{-1}\frac{1-x}{1+x} ,\ x\in [ 0,1]\\
Solution:\ RHS=\frac{1}{2} cos^{-1}\frac{1-x}{1+x}\\
put\ x=tan^{2} \theta \Longrightarrow tan\theta =\sqrt{x}\\
\Longrightarrow \theta =tan^{-1}\sqrt{x}\\
=\frac{1}{2} cos^{-1}\left(\frac{1-tan^{2} \theta }{1+tan^{2} \theta }\right)\\
=\frac{1}{2} cos^{-1}( cos2\theta )\\
=\frac{1}{2} 2\theta =\theta \\
=tan^{-1}\sqrt{x}
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Q.9.NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise

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Q.9.\ Prove\ that:\ \\
cot^{-1}\left(\frac{\sqrt{1+sinx} +\sqrt{1-sinx}}{\sqrt{1+sinx} -\sqrt{1-sinx}}\right) =\frac{x}{2} ,\ x\in \left( 0,\frac{\pi }{4}\right)\\
Solution:\ 1+sinx=cos^{2}\frac{x}{2} +sin^{2}\frac{x}{2} +2cos\frac{x}{2} sin\frac{x}{2}\\
\Longrightarrow \sqrt{1+sinx} =cos\frac{x}{2} +sin\frac{x}{2}\\
\Longrightarrow \sqrt{1-sinx} =cos\frac{x}{2} -sin\frac{x}{2}\\
Substituting\ the\ above\ identities,\\
cot^{-1}\left(\frac{cos\frac{x}{2} +sin\frac{x}{2} +cos\frac{x}{2} -sin\frac{x}{2}}{cos\frac{x}{2} +sin\frac{x}{2} -cos\frac{x}{2} +sin\frac{x}{2}}\right)\\
=cot^{-1}\left(\frac{2cos\frac{x}{2}}{2sin\frac{x}{2}}\right) =cot^{-1}\left( cot\frac{x}{2}\right)\\
=\frac{x}{2} \ Ans.
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Q.10. NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise

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Q.\ 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\\
Solution:\ LHS=tan^{-1}\left(\frac{\sqrt{1+x} -\sqrt{1-x}}{\sqrt{1+x} +\sqrt{1-x}}\right)\\
Put\ x=cos\theta \\
tan^{-1}\left(\frac{\sqrt{1+cos\theta } -\sqrt{1-cos\theta }}{\sqrt{1+cos\theta } +\sqrt{1-cos\theta }}\right)\\
=tan^{-1}\left(\frac{\sqrt{2cos^{2}\frac{\theta }{2}} -\sqrt{2sin^{2}\frac{\theta }{2}}}{\sqrt{2cos^{2}\frac{\theta }{2}} +\sqrt{2sin^{2}\frac{\theta }{2}}}\right)\\
\ [ using\ identity,\ 1+cos\theta =2cos^{2}\frac{\theta }{2} \ and\ 1-cos\theta =2sin^{2}\frac{\theta }{2}\\
=tan^{-1}\left(\frac{cos\frac{\theta }{2} -sin\frac{\theta }{2}}{cos\frac{\theta }{2} +sin\frac{\theta }{2}}\right)\\
=tan^{-1}\left(\frac{\frac{cos\frac{\theta }{2}}{cos\frac{\theta }{2}} -\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}}{\frac{cos\frac{\theta }{2}}{cos\frac{\theta }{2}} +\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}}\right) \ \left[ Dividin\ numerator\ and\ denominator\ by\ cos\frac{\theta }{2}\right]\\
=tan^{-1}\left(\frac{1-tan\frac{\theta }{2}}{1+tan\frac{\theta }{2}}\right)\\
=tan^{-1}\left(\frac{tan\frac{\pi }{4} -tan\frac{\theta }{2}}{1+tan\frac{\pi }{4} .tan\frac{\theta }{2}}\right)\\
=tan^{-1}\left( tan\left(\frac{\pi }{4} -\frac{\theta }{2}\right)\right)\\
=\frac{\pi }{4} -\frac{\theta }{2} =\frac{\pi }{4} -\frac{1}{2} cos^{-1} x\ Proved.\\
\
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Q. 11. NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution

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\ Solve\ the\ following\ equations:\\
Q.11.\ 2tan^{-1}( cosx) =tan^{-1}( 2cosecx)\\
Solution:\ using\ identity\ 2tan^{-1} x=tan^{-1}\left(\frac{2x}{1-x^{2}}\right)\\
tan^{-1}\left(\frac{2cosx}{1-cos^{2} x}\right) =tan^{-1}( 2cosecx)\\
\Longrightarrow \left(\frac{2cosx}{sin^{2} x}\right) =2cosecx\\
\Longrightarrow 2cosecxcotx=2cosecx\\
\Longrightarrow cotx=1\\
\Longrightarrow x=\frac{\pi }{4}\\
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Q. 12. NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution

Q. 13. NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution

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Q.\ 13.\
sin\left( tan^{-1} x\right) ,\ |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^{-1} x=\alpha \\
\Longrightarrow tan\alpha =x\Longrightarrow sin\alpha =\frac{x}{\sqrt{1+x^{2}}}\\
\Longrightarrow \alpha =sin^{-1}\frac{x}{\sqrt{1+x^{2}}}\\
\Longrightarrow sin\left( sin^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)\right) =\frac{x}{\sqrt{1+x^{2}}} \ Ans.\ ( D)
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Q.14.NCERT Class 12 Maths Chapter 2 Miscellaneous Exercise Solution

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Q.14.\ sin^{-1}( 1-x) -2sin^{-1} x=\frac{\pi }{2}\\
Put\ x=sin\theta \\
sin^{-1}( 1-sin\theta ) -2sin^{-1}( sin\theta ) =\frac{\pi }{2}\\
\Longrightarrow sin^{-1}( 1-sin\theta ) -2\theta =\frac{\pi }{2}\\
\Longrightarrow sin^{-1}( 1-sin\theta ) =\frac{\pi }{2} +2\theta \\
\Longrightarrow ( 1-sin\theta ) =sin\left(\frac{\pi }{2} +2\theta \right)\\
\Longrightarrow 1-sin\theta =cos2\theta \\
\Longrightarrow 1-sin\theta =1-2sin^{2} \theta \\
\Longrightarrow 2sin^{2} \theta -sin\theta =0\\
\Longrightarrow sin\theta ( 2sin\theta -1) =0\\
Back\ substituting\ sin\theta =x\\
\Longrightarrow x( 2x-1) =0\\
\Longrightarrow x=0\ or\ x=\frac{1}{2}\\
Ans.\ ( A)
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Conclusion

The NCERT Class 12 Maths , Chapter 2 Miscellaneous Exercise Solution, is a powerful resource for students who want to score high in their board exams. It will strengthen their foundational knowledge and boost their problem-solving skills. This is especially important for students who are self-studying and do not have access to a teacher’s support. In such cases, the NCERT Class 12 Maths Chapter 2 (Inverse Trigonometry) Miscellaneous Exercise Solutions will prove to be very helpful. We have provided this solution step by step.