Prove that (secA - tanA)^2 = 1-sinA/1+sinA
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Answer:
1-sinA/1+sinA=(secA-tanA)²
L.H.S
1-sinA/1+sinA
Rationalising the denominator
1-sinA (1-SinA) /1+sinA(1-SinA)
(1-SinA)²/ 1² -(Sin²A)
(1-SinA)²/ 1 -(Sin²A)
(1-SinA)² /Cos² A
[ 1 -Sin²A = cos²A]
(1-SinA/CosA)²
(1/CosA-SinA/CosA)²
(SecA-tanA)²
L.H S = R H.S
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