Question

1+secA/secA=sinsquareA/1- cosA

Answer 1

$\mathfrak{\huge{\green{\underline{\underline{ANSWER :}}}}}$

LHS = ( 1 + secA )/secA

= ( 1 + 1/cosA ) / ( 1/cosA )

= [ ( CosA + 1 ) / cosA ] / ( 1/ cosA )

= Cos A + 1

= ( 1 + cosA ) ( 1 - cosA ) / ( 1 - cosA )

= ( 1 - cos² A ) / ( 1 - cosA )

= Sin² A / ( 1 - cosA )

= RHS

Hope it will help you.

Answer 2

$\huge{\underline{\mathbb{\red{Answer}}}}$

LHS = ( 1 + secA )/secA

= ( 1 + 1/cosA ) / ( 1/cosA )

= [ ( CosA + 1 ) / cosA ] / ( 1/ cosA )

= Cos A + 1

= ( 1 + cosA ) ( 1 - cosA ) / ( 1 - cosA )

= ( 1 - cos² A ) / ( 1 - cosA )

= Sin² A / ( 1 - cosA )

= RHS

$\huge{\red{\ddot{\smile}}}$

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$\huge{\underline{\underline{\mathfrak{Thank you}}}}$