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If \( A+B=90^{\circ} \), prove that\[\frac{\cos A}{1+\cos B}+\frac{1+\cos B}{\cos A}=\frac{2}{\cos A}\] |
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Answer» \(LHS\) = \(\frac{\cos A}{1+\cos B}+\frac{1+\cos B}{\cos A}\) \(LHS\) \(=\) \(\frac{cos^2A + (1 + cosB)^2}{cosA + cosA.cosB}\) \(LHS\) \(=\) \(\frac{cos^2A + 1 + cos^2B + 2cosB}{cosA + cosA.cosB}\) \(LHS\) \(=\) \(\frac{cos^2A + 1 + sin^2A + 2cosB}{cosA + cosA.cosB}\) \([Cos(90 - θ) = Sinθ]\) \(LHS\) \(=\) \(\frac{2 + 2cosB}{cosA + cosA.cosB}\) \([sin^2A + cos^2A = 1]\) \(LHS\) \(=\) \(\frac{2(1 + cosB)}{cosA (1 + cosB)}\) \(LHS\) \(=\frac{2}{\cos A} \) = \(RHS \) Hence, proved |
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