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Mathematics Question on Inverse Trigonometric Functions

Write the function in the simplest form: tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x)0<x<π\tan^{-1}\bigg(\frac{\cos x-\sin x}{\cos x+\sin x}\bigg)0<x<\pi

Answer

tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x)\tan^{-1}\bigg(\frac{\cos x-\sin x}{\cos x+\sin x}\bigg)

= tan⁡−1(1−sin⁡xcos⁡x1+sin⁡xcos⁡x)\tan^{-1}\bigg(\frac{\frac{1-\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}\bigg)

=tan⁡−1(1−tan⁡x1+tan⁡x)\tan^{-1}\bigg(\frac{1-\tan x}{1+\tan x}\bigg)

=tan⁡−1(1)−tan⁡−1(tan⁡x) [tan⁡−1x−y1−xy=tan⁡−1x−tan⁡−1y]\tan^{-1}(1) -\tan^{-1}(\tan x)\:[\tan^{-1}\frac{x-y}{1-xy}=\tan^{-1}x-\tan^{-1}y]

= π4−x\frac{\pi}{4-x}