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Math Help - Prove: Trigonometric Identities

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    Prove: Trigonometric Identities

    [1+tan(x/2)] / [1-tan(x/2)] = sec x + tan x
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    Quote Originally Posted by Quan View Post
    [1+tan(x/2)] / [1-tan(x/2)] = sec x + tan x
    \frac{1+\tan \frac{x}{2}}{1 - \tan \frac{x}{2}} \cdot \frac{\cos \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2} }{\cos \frac{x}{2} - \sin \frac{x}{2}}

    \frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \cdot \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} = \frac{cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2\sin \frac{x}{2}\cos \frac{x}{2}}{\cos^2 \frac{x}{2} - \sin \frac{x}{2}}

    \frac{1 + \sin x}{\cos x} = \sec x + \tan x
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