Prove that √1+sinA/1-sinA=secA+tanA
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Answer:
LHS = √(1 + sin∅)/√(1 - sin∅)
= √(1 + sin∅) × √(1 + sin∅)/√(1 -sin∅)×√(1 +sin∅)
= √(1 + sin∅)²/√(1 -sin²∅)
= (1 + sin∅)/√cos²∅
= (1 + sin∅)/cos∅
= 1/cos∅ + sin∅/cos∅
= sec∅ + tan∅ = RHS
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