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Question Number 191477 by Spillover last updated on 24/Apr/23

$$\mathrm{Prove}\mathrm{that}$$$$$$$$\frac{\tan \mathrm{y}+\sec \mathrm{y}-1}{\tan \mathrm{y}-\sec \mathrm{y}+1}=\tan \mathrm{y}+\sec \mathrm{y}$$$$$$

Answered by ARUNG_Brandon_MBU last updated on 24/Apr/23

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

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