Question

prove the following​

Answer 1

$\sf{\orange{\underline{\huge{\mathsf{Solution}}}}}$

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$\sf: \implies\: {\bold{ \dfrac{CosA}{1 - tanA} + \dfrac{sin^2A}{SinA - CosA} = Sin A + CosA}}$

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$\sf{\red{\boxed{\bold{tanA = \dfrac{SinA}{CosA}}}}}$

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$\sf: \implies\: {\bold{ \dfrac{CosA}{1 - \dfrac{sinA}{CosA}} + \dfrac{sin^2A}{SinA - CosA} = Sin A + CosA}}$

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$\sf: \implies\: {\bold{ \dfrac{CosA}{\dfrac{cosA - sinA}{CosA}} + \dfrac{sin^2A}{SinA - CosA} = Sin A + CosA}}$

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$\sf: \implies\: {\bold{ {CosA}\div \dfrac{cosA - sinA}{CosA} + \dfrac{sin^2A}{SinA - CosA} = Sin A + CosA}}$

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$\sf: \implies\: {\bold{{CosA}\times \dfrac{CosA}{CosA - SinA} + \dfrac{sin^2A}{SinA - CosA} = Sin A + CosA}}$

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$\sf: \implies\: {\bold{ \dfrac{Cos^2A}{CosA - SinA } + \dfrac{sin^2A}{SinA - CosA} = Sin A + CosA}}$

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$\sf: \implies\: {\bold{ \dfrac{Cos^2A}{- ( SinA - CosA )} + \dfrac{sin^2A}{SinA - CosA} = Sin A + CosA}}$

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$\sf: \implies\: {\bold{ \dfrac{ - Cos^2A}{SinA - CosA } + \dfrac{sin^2A}{SinA - CosA} = Sin A + CosA}}$

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$\sf: \implies\: {\bold{ \dfrac{-Cos^2A + Sin^2A}{SinA - CosA} = SinA + CosA}}$

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$\sf: \implies\: {\bold{ \dfrac{Sin^2A - Cos^2A}{SinA - CosA} = SinA + CosA}}$

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$\sf{\red{\boxed{\bold{A^2 - B^2=( A + B) ( A - B)}}}}$

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$\sf: \implies\: {\bold{ \dfrac{( sinA - CosA)( sinA + CosA)}{SinA - CosA} = SinA + CosA}}$

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$\sf: \implies\: {\bold{ \dfrac{( sinA - CosA)\cancel{( sinA + CosA)}}{\cancel{SinA - CosA}} = SinA + CosA}}$

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$\sf: \implies\: {\bold{ SinA + CosA = SinA + CosA}}$

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⠀⠀⠀⠀⠀⠀⠀⠀.....Hence Proved

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Answer 2

$\tt \color{green} \large \underline{ \underline{ Solution}} \\ \\ \\ \\ \tt \: L. H. S \\ \\ \tt \frac{cosA}{ \frac{1 - sinA}{cosA} } + \frac{ {sin}^{2}A }{sinA - cosA} \\ \\ \\ \\ \tt : \implies \frac{cosA}{ \frac{cosA - sinA}{cosA} } + \frac{ {sin}^{2} A}{sinA - cosA} \\ \\ \\ \\ \tt : \implies \frac{cosA \times cosA}{cosA - sinA} - \frac{ {sin}^{2}A }{cosA - sinA} \\ \\ \\ \\ : \implies \tt \frac{ {cos}^{2}A }{cosA - sinA } - \frac{ {sin}^{2} A}{cosA - sinA} \\ \\ \\ \\ \tt : \implies \frac{ {cos}^{2} A - {sin}^{2} A}{ \big(cosA - sinA \big)} \\ \\ \\ \\ \tt : \implies \frac{ \cancel{\big(cosA - sinA \big) } \big(cosA + sinA \big)}{ \cancel{cosA - sinA}} \\ \\ \\ \\ \tt : \implies cosA + sinA \\ \\ \\ \\ \tt \color{green}L. H. S = R. H. S$