Question

Prove that
$\sqrt{ \frac{tano - coto \: }{sino coso} = {tano}^{2} - {coto }^{2}$
​

Answer 1

Answer:

I didn't understand anything

Answer 2

LHS.

$\frac{ \tan( \alpha ) - \cot( \alpha ) }{ \sin( \alpha ) \cos( \alpha ) } = \frac{ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } }{ \sin( \alpha ) \cos( \alpha ) } =$

$\frac{ \sin {}^{2} ( \alpha ) - \cos {}^{2} ( \alpha ) }{ \sin {}^{2} ( \alpha ) \cos {}^{2} ( \alpha ) } = \frac{1}{ \cos {}^{2} ( \alpha ) } - \frac{1}{ \sin {}^{2} ( \alpha ) }$

$= \sec {}^{2} ( \alpha ) - \csc {}^{2} ( \alpha ) = 1 + \tan {}^{2} ( \alpha ) - 1 - \cot {}^{2} ( \alpha ) =$

$\tan {}^{2} ( \alpha ) - \cot {}^{2} ( \alpha )$

RHS.

PROVED.

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