Question

tan⁻¹(√1+cosx+√1-cosx/√1+cosx-√1-cosx) = π/4 -x/2, π < x < 3π/2,prove it

Answer 1

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