Question

Prove that
$\cos( \alpha ) . \sin(90 - \alpha ) - \frac{ \sin( \alpha ) \cos(90 - \alpha ) \cos( \alpha ) }{ \sec(90 - \alpha ) } - \frac{ \cos( \alpha ) \sin(90 - \alpha ) \sin( \alpha ) }{ \csc(90 - \alpha ) } + \csc(90 - \alpha ) = \frac{1}{ \cos( \alpha ) }$
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Answer 1

Step-by-step explanation:

cos( alpha ) = sin( 90 - alpha )

sin( alpha ) = cos( 90 - alpha )

sec(90 - alpha ) = cosec ( alpha ) = 1/sin( alpha )

csc (90 - alpha ) = sin ( alpha)

Put these values