Prove that root sec theta minus one upon sec theta + 1 +
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√secA-1/√secA+1 +√secA+1/√secA-1 = 2cosec A
LHS = (√secA-1)² + (√secA+1)² / √sec²A-1
= secA-1 + secA+1 / √tan²A
= 2SecA/tanA
= [2×(1/cos)]/(sinA/cosA)
= 2×(1/cos)×(cosA/sinA)
= 2 × 1/sinA
= 2 CosecA
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