Question

(secA-tanA)^2=1-sinA/1+sinA

Answer 1

Solution==>>

solving Right hand side

(1-sinA)/(1+sinA)

=(1-sinA)(1-sinA)/(1+sinA)(1-sinA)

=(1-sinA)²/(1-sin²A)

=(1-sinA)²/cos²A

={(1-sinA)/cosA}²

={(1/cosA)-(sinA/cosA)}²

={secA-tanA}²..........proved

Answer 2

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