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

Prove that

\frac{\tan y + \sec y - 1}{\tan y - \sec y + 1} = \tan y + \sec 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