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#### Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 4 maths Textbook Solution.

Answer: $\mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$

Given: $2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x), x \neq \frac{\pi}{2}$

Hint: Using $2 \tan ^{-1}(x)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

Solution: We have,

$2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x)$

We know that

$2 \tan ^{-1}(x)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{2 \sin x}{1-\sin ^{2} x}\right)=\tan ^{-1}(2 \sec x) \\ &\Rightarrow \frac{2 \sin x}{\cos ^{2} x}=2 \sec x \end{aligned}                                    $\left[1-\sin ^{2} x=\cos ^{2} x\right]$

$\Rightarrow \frac{\sin x}{\cos x \cos x}=\frac{1}{\cos x}$                                                                    $\left[\sec x=\frac{1}{\cos x}\right]$

$\Rightarrow \tan x=1$                                                                                       $\left[\frac{\sin x}{\cos x}=\tan x\right]$

\begin{aligned} &\Rightarrow x=\tan ^{-1}(1) \\ &\Rightarrow x=\frac{\pi}{4} \end{aligned}                                                                                $\left[\tan ^{-1}(1)=\frac{\pi}{4}\right]$

Hence the principal value of $x=n \pi+\frac{\pi}{4}, n \in Z$