√1-sinA/√1+sinA = secA-tanA
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Step-by-step explanation:
LHS = √(1-sinA) ÷ √(1+sinA)
rationalising
=> √{ (1-sinA) × (1-sinA) } ÷ √{ (1+sinA) (1-sinA)}
using (a²-b²) = (a+b)(a-b)
=> √ { (1-sinA)²} ÷ √ (1 - sin²A)
=> (1-sinA) ÷ √ ( cos²A)
=> (1-sinA) ÷ (cosA)
=> (1/cosA) - (sinA/cosA)
=> secA - tanA = RHS
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