prove that 1-sin A upon 1+si A = ( sec A - tan A ) 2
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Answer:
√(1+sinA)/(1-sinA) × (1+sinA)/(1+sinA) = √(1+sinA)^2/1-sin^2A
√(1+sinA)^2/cos^2A =1+sinA)cosA
1/cosA + sinA/cosA= secA+tanA
Hence proved
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Answer:
l think in rhs after (secA-tanA) you tried to write square of this
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