prove that √1-sinA/1+sinA=secA-tanA
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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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