Question:
Prove that (1 + sec A)/sec A = sin2 A/(1 - cos A)
Answer:
LHS:
(1 + sec A)/sec A
= (1 + 1/cos A)/(1/cos A)
= {(cos A + 1)/cos A}/(1/cos A)
= {(cos A + 1) * cos A}/(cos A/1)
= cos A + 1
RHS:
sin2 A/(1 - cos A)
= (1 - cos2 A)/(1 - cos A) {since sin2 A + cos2 A = 1}
= {(1 - cos A) * (1 + cos A)}/(1 - cos A)
= 1 + cos A
Since, LHS = RHS = 1 + cos A
So, (1 + sec A)/sec A = sin2 A/(1 - cos A)
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