Prove that .1+ sec A/secA=
sin^2A/1-cosA
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Answered by
154
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
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Step-by-step explanation:
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