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- tan A cotA20. Prove that —— .+ =~e 1-cotA 1-tanA =1+ SecAcosec A |
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Answer» LHS:tan A/(1 - cot A) + cot A /(1 - tan A) = (tan A)/[(1 - (1/tan A)] + cot A /(1 - tan A) = (tan^2 A)/[(tan A - 1)] + cot A /(1 - tan A) = (tan^2 A)/[(tan A - 1)] - cot A /(tan A - 1) = (tan^2 A - cot A) / (tan A - 1) = (tan^2 A - 1/tan A) / (tan A - 1) = (tan^3 A - 1) / [tan A (tan A - 1)] = (tan A - 1)(tan^2 A + tan A + 1) / [tan A (tan A - 1)] = (tan^2 A + tan A + 1) / tan A = 1 + tan A + cot A = 1 + [(sin A/cosA) + (cos A/sin A)] = 1 + [(sin^2 A + cos^2 A) / sin A cos A] = 1 + [1 / (sin A cos A)] = 1 + (sec A x cos A)= RHS Hence proved |
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