Question

prove the following identify, where the angels involved are acute angles for which the expressions are defined. tan A/1-cot A+cot A/1-tan A=1+sec A cosec A.
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Answer 1

$\tt To\:prove : \frac{tanA}{1-cotA} + \frac{cotA}{1-tanA} = 1+ secA.coSecA \\ \\ \tt L.H.S \\ \\ \tt \frac{tanA}{1-cotA} + \frac{cotA}{1-tanA} \\ \\ \tt Converting\:into\:sinA\:and\:cosA \\ \\ \tt \implies \frac{\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}} + \frac{\frac{cosA}{sinA}}{1-\frac{sinA}{cosA}} \\ \\ \tt \implies \frac{\frac{sinA}{cosA}}{\frac{sinA - cosA}{sinA}} + \frac{\frac{cosA}{sinA}}{\frac{cosA - sinA}{cosA}} \\ \\ \tt \implies \frac{{sin}^{2}A}{cosA(sinA-cosA)}. + \frac{{cos}^{2}A}{sinA(cosA-sinA)} \\ \\ \tt \implies \frac{{sin}^{2}A}{cosA(sinA-cosA)} - \frac{{cos}^{2}A}{sinA(sinA -cosA)} \\ \\ \tt \implies \frac{{sin}^{3}A}{sinA.cosA(sinA-cosA)} - \frac{{cos}^{3}A}{sinA.cosA(sinA-cosA)} \\ \\ \tt \implies \frac{{sin}^{3}A - {cos}^{3}A}{sinA.cosA(sinA-cosA)} \\ \\ \tt \implies \frac{(sinA-cosA)({sin}^{2}A+{cos}^{2}A+sinA.cosA)}{sinA.cosA.(sinA - cosA)} \\ \\ \tt \implies \frac{\orange{\cancel{(sinA-cosA)}}(\green{{sin}^{2}A + {cos}^{2}A} + sinA.cosA}{sinA.cosA \orange{ \cancel{(sinA-cosA)}}} \\ \\ \tt \implies \frac{1+sinA.cosA}{sinA.cosA} \\ \\ \tt \implies \frac{1}{sinA.cosA} + \frac{sinA.cosA}{sinA.cosA} \\ \\ \tt \implies 1 + secA.cosecA \\ \\ \tt \implies R.H.S \\ \\ \tt \underline{\red{\tt Hence\:proved}}$

Answer 2

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