Question

Evaluate: ${\tan }^{-1}\sqrt{\frac{\left(1-\sin 4x\right)}{\left(1+\sin 4x\right)}}$

Answer 1

$ta{n}^{-1}\sqrt{\frac{(1-\sin 4x)}{(1+\sin 4x)}}$

$=ta{n}^{-1}\sqrt{\frac{(si{n}^{2}2x+{\cos }^{2}2x-2\sin 2x\cos 2x)}{(si{n}^{2}2x+{\cos }^{2}2x+2\sin 2x\cos 2x)}}$

$={\tan }^{-1}\sqrt{\frac{{(\sin 2x-\cos 2x)}^2}{{(\sin 2x+\cos 2x)}^2}}$

$={\tan }^{-1}\frac{(\sin 2x-\cos 2x)}{(\sin 2x+\cos 2x)}$

Divide numerator and denominator $\cos 2x$ and we get,

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

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

$={\tan }^{-1}\left(\frac{\tan 2x-\tan {45}^0}{\tan 2x+1}\right)$

$={\tan }^{-1}\left(\frac{\tan 2x-\tan {45}^0}{1+1\times \tan 2x}\right)$

$={\tan }^{-1}\left(\frac{\tan 2x-\tan {45}^0}{1+\tan {45}^0\tan 2x}\right)$

$={\tan }^{-1}\tan (2x-{45}^0)$

$=2x-{45}^{0}or2x-\frac{\pi }{4}$

Hence, this is the solution.