If x∈−π2,π2 , then the value of tan−1⁡tan⁡x4+tan−1⁡3sin⁡2x5+3cos⁡2x is

tan−1⁡tan⁡x4+tan−1⁡3sin⁡2x5+3cos⁡2x=tan−1⁡tan⁡x4+tan−1⁡6tan⁡x1+tan2⁡x5+31−tan2⁡x1+tan2⁡x=tan−1⁡tan⁡x4+tan−1⁡6tan⁡x8+2tan2⁡x=tan−1⁡tan⁡x4+tan−1⁡3tan⁡x4+tan2⁡x=tan−1⁡tan⁡x4+3tan⁡x4+tan2⁡x1−3tan2⁡x44+tan2⁡xas⁡tan⁡x4⋅3tan⁡x4tan2⁡x<1=tan−1⁡16tan⁡x+4tan3⁡x16+4tan2⁡x=tan−1⁡(tan⁡x)=x

If x∈−π2,π2 , then the value of tan−1⁡tan⁡x4+tan−1⁡3sin⁡2x5+3cos⁡2x is

tan−1⁡tan⁡x4+tan−1⁡3sin⁡2x5+3cos⁡2x=tan−1⁡tan⁡x4+tan−1⁡6tan⁡x1+tan2⁡x5+31−tan2⁡x1+tan2⁡x=tan−1⁡tan⁡x4+tan−1⁡6tan⁡x8+2tan2⁡x=tan−1⁡tan⁡x4+tan−1⁡3tan⁡x4+tan2⁡x=tan−1⁡tan⁡x4+3tan⁡x4+tan2⁡x1−3tan2⁡x44+tan2⁡xas⁡tan⁡x4⋅3tan⁡x4tan2⁡x<1=tan−1⁡16tan⁡x+4tan3⁡x16+4tan2⁡x=tan−1⁡(tan⁡x)=x

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If $x \in (- \frac{π}{2}, \frac{π}{2})$ , then the value of $\tan^{- 1} (\frac{\tan x}{4}) + \tan^{- 1} (\frac{3 \sin 2 x}{5 + 3 \cos 2 x})$ is

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$\frac{x}{2}$

answer is D.

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Detailed Solution

$\begin{matrix} \tan^{- 1} (\frac{\tan x}{4}) + \tan^{- 1} (\frac{3 \sin 2 x}{5 + 3 \cos 2 x}) \\ = \tan^{- 1} (\frac{\tan x}{4}) + \tan^{- 1} (\frac{\frac{6 \tan x}{1 + \tan^{2} x}}{5 + \frac{3 (1 - \tan^{2} x)}{1 + \tan^{2} x}}) \\ = \tan^{- 1} (\frac{\tan x}{4}) + \tan^{- 1} (\frac{6 \tan x}{8 + 2 \tan^{2} x}) \\ = \tan^{- 1} (\frac{\tan x}{4}) + \tan^{- 1} (\frac{3 \tan x}{4 + \tan^{2} x}) \\ = \tan^{- 1} (\frac{\frac{\tan x}{4} + \frac{3 \tan x}{4 + \tan^{2} x}}{1 - \frac{3 \tan^{2} x}{4 (4 + \tan^{2} x)}}) (as |\frac{\tan x}{4} \cdot \frac{3 \tan x}{4 \tan^{2} x}| < 1) \\ = \tan^{- 1} (\frac{16 \tan x + 4 \tan^{3} x}{16 + 4 \tan^{2} x}) = \tan^{- 1} (\tan x) = x \end{matrix}$