Question

prove the following

Answer 1

last one , LHS :
$\sin(90 - \alpha ) - \cos(90 - \alpha ) = \sin( \alpha ) \times \cos( \alpha )$
Rhs
$\tan( \alpha ) \div1 + \tan( \alpha ) = \tan( \alpha ) \div \sec^{2} ( \alpha ) = \sin( \alpha ) \times \cos^{2} ( \alpha ) \div { \cos( \alpha ) } = \sin( \alpha ) \times \cos( \alpha )$
hence lhs=rhs ,hence proved.