1.

PROVE THE IDENTITIES A) sin A/1+cos A≡1-cos A/sin A B) cos A/1- tan A + sin A/1-cot A ≡sin A + cos A

Answer»

Step-by-step EXPLANATION:

A)

LHS =

\frac{ \sin( \alpha ) }{1 + \cos( \alpha ) }

= \frac{ \sin( \alpha ).(1 - \cos( \alpha ) }{(1 + \cos( \alpha ) ).(1 - \cos( \alpha )) }

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

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

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

= RHS

Hence PROVED

B)

LHS =

\frac{ \cos( \alpha ) }{1 - \tan( \alpha ) } + \frac{ \sin( \alpha ) }{1 - \cot( \alpha ) }

= \frac{ \cos( \alpha ) }{1 - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } + \frac{ \sin( \alpha ) }{1 - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } }

= \frac{ \cos( \alpha ) \cos( \alpha ) }{ \cos( \alpha )(1 - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } ) } + \frac{ \sin( \alpha ) \sin( \alpha ) }{ \sin( \alpha )(1 - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } ) }

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

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

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

= \frac{( \cos( \alpha ) + \sin( \alpha )) ( \cos( \alpha ) - \sin( \alpha )) }{\cos( \alpha ) - \sin( \alpha )) }

= \sin( \alpha ) + \cos( \alpha )

= RHS

Hence Proved


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