Question

tan( pie/4+ x/2)= (1+sinx/1-sinx)^1/2

Answer 1

Answer:

RHS

$\sqrt{ \frac{1 + sin \: x}{1 - sin \: x} } = \sqrt{ \frac{ \cos {}^{2} ( \frac{x}{2} ) + \sin {}^{2} ( \frac{x}{2} ) + 2 \sin( \frac{x}{2} ) \cos( \frac{x}{2} ) }{ \cos {}^{2} ( \frac{x}{2} ) + \sin {}^{2} ( \frac{x}{2} ) - 2 \sin( \frac{x}{2} ) \cos( \frac{x}{2} ) } } =$

$\sqrt{ \frac{( \cos( \frac{x}{2} ) + \sin( \frac{x}{2} )) {}^{2} }{ (\cos( \frac{x}{2} ) - \sin( \frac{x}{2} )) {}^{2} } } = \frac{ \cos( \frac{x}{2} ) + \sin( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) - \sin( \frac{x}{2} ) } =$

now dividing nr. & dn. by cos(x/2)...

$\frac{1 + \tan( \frac{x}{2} ) }{1 - \tan( \frac{x}{2} ) } = \frac{ \tan( \frac{\pi}{4} ) + \tan( \frac{x}{2} ) }{1 - \tan( \frac{\pi}{4} ) \tan( \frac{x}{2} ) } =$

$\tan( \frac{\pi}{4} + \frac{x}{2} ) \: \: \: proved$

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