Question

Guys please answer this question quickly.I need it right now. I will mark the best answer as brainliest.​

Answer 1

$\huge \underline \bold \green{QUESTION}:$

$\frac{ \sqrt{1 + cosx } +\sqrt{1 - cosx } }{\sqrt{1 + cosx } - \sqrt{1 - cosx }}$

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$\huge \underline \bold \red{SOLUTION}:$

$\frac{ \sqrt{1 + cosx } +\sqrt{1 - cosx } }{\sqrt{1 + cosx } - \sqrt{1 - cosx }} \\ = > \frac{ \sqrt{2 cos {}^{2} (\frac{x}{2}) } + \sqrt{2 sin {}^{2} (\frac{x}{2}) }}{ \sqrt{2 cos {}^{2} (\frac{x}{2}) } - \sqrt{2 sin {}^{2} (\frac{x}{2}) } } \\ = > \frac{ \cos( \frac{x}{2} ) + \sin( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) - \sin( \frac{x}{2} )} \\ \red{divide \: whole \: equn \: by \: cos \frac{x}{2} } \\ = > \frac{1 + \tan( \frac{x}{2} ) }{1 - \tan( \frac{x}{2} )} \\ \green{\boxed{= > \tan( \frac{\pi}{4} + \frac{x}{2} ) }}$

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$\underline\bold \red{final \: answer} :$

$\green{\boxed{ \orange{ \tan( \frac{\pi}{4} + \frac{x}{2} ) }}}$

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$\underline\bold \red{formulas \: used} :$

$= > 1 + \cos( \alpha ) = 2 { cos }^{2} \frac{ \alpha }{2} \\ = > 1 - \cos( \alpha ) = 2 \sin {}^{2} \frac{ \alpha }{2} \\ = > \frac{1 + \tan( \alpha ) }{1 - \tan( \alpha ) } = \tan( \frac{\pi}{4} + \alpha )$

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hope it helps you