Question

it's urgent plz solve this
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Answer 1

see the attachement

Khushi ✔️✌

Answer 2

RHS

${( \sec( \alpha ) - \tan( \alpha ) })^{2} \\ {( \frac{1 - \sin( \alpha ) }{ \cos( \alpha ) } })^{2} \\ we \: know \: that \\ 1 - { \sin( \alpha ) }^{2} = { \cos( \alpha ) }^{2} \\ so \\ \frac{(1 - \sin( \alpha )(1 - \sin( \alpha ) ) }{(1 - \sin( \alpha )(1 + \sin( \alpha ) ) } \\ \frac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) }$

Hence RHS = LHS