Question

Prove that tan ^-1(√1+cosx+√1-cosx/√1+cosx-√1-cosx) = π/4-x/2

Answer 1

you can see your answer

Answer 2

$\tan^{-1}[\frac{\sqrt{1 + \cos x} + \sqrt{1 - \cos x}}{\sqrt{1 + \cos x} - \sqrt{1 - \cos x}}] \\ \\ = \tan^{-1}[\frac{\sqrt{1 + 2 \cos^{2}\frac{x}{2} -1 } + \sqrt{1 - 1 + 2 \sin^{2}\frac{x}{2} }}{\sqrt{1 + 2 \cos^{2}\frac{x}{2} -1 } - \sqrt{1 - 1 + 2 \sin^{2}\frac{x}{2}} }] \\ \\= \tan^{-1}(\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}) \\ Since, \ \pi<x < 3\pi/2 \\ \\ Therefore, \ sin\frac{x}{2} \ \ \text{Will be negative} \\ \\ = \tan^{-1}(\frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}) \\ \\= \tan^{-1}(\frac{1 - \tan \frac{x}{2}}{1 + \tan\frac{x}{2}}) \\ \\ = \tan^{-1}(\frac{\tan\frac{\pi}{4} - tan \frac{x}{2}}{\tan\frac{\pi}{2} + \sin\frac{x}{2}}) \\ \\ = \tan^{-1}(\tan\frac{\pi}{4}) - \tan^{-1}(\tan\frac{x}{2}) \\ \\ = \frac{\pi}{4} - \frac{x}{2}$