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Mathematics Question on Inverse Trigonometric Functions

If x∈(−π2,π2)x \in \left(-\frac{\pi}{2}, \frac{\pi }{2}\right), then the value of tan−1(tan x4)+tan−1(3 sin 2x5+3 cos 2x)tan^{-1}\left(\frac{tan\,x}{4}\right)+tan^{-1}\left(\frac{3\,sin\,2x}{5 + 3\,cos\,2x}\right) is

A

x2\frac{x}{2}

B

2x2x

C

3x3x

D

xx

Answer

xx

Explanation

Solution

tan−1(tan x4)+tan−1(3 sin 2x5+3 cos 2x)tan^{-1}\left(\frac{tan\,x}{4}\right)+tan^{-1}\left(\frac{3\,sin\,2x}{5 + 3\,cos\,2x}\right) =tan−1(tan x4)+tan−1(6 tan x1+tan2 x5+3(1−tan2 x1+tan2 x))= tan^{-1}\left(\frac{tan\,x}{4}\right)+tan^{-1}\left(\frac{\frac{6\,tan\,x}{1+tan^{2}\,x}}{5 + 3\left(\frac{1-tan^{2}\,x}{1+tan^{2}\,x}\right)}\right) =tan−1(tan x4)+tan−1(3 tan x4+tan2 x)= tan^{-1}\left(\frac{tan\,x}{4}\right) + tan^{-1}\left(\frac{3\,tan\,x}{4+tan^{2}\,x}\right) =tan−1(tan x4+3 tan x4+tan2 x1−3 tan x4(4+tan2 x))= tan^{-1}\left(\frac{\frac{tan\,x}{4}+\frac{3\,tan\,x}{4+tan^{2}\,x}}{1-\frac{3\,tan\,x}{4\left(4+tan^{2}\,x\right)}}\right) =tan−1(16 tan x+tan3 x16+tan2 x)= tan^{-1}\left(\frac{16\,tan\,x+tan^{3}\,x}{16 + tan^{2}\,x}\right) =tan−1(tanx)=x= tan^{-1}\left(tanx\right) = x