Question

prove that 1+cosA-sin2A/sinA(1+cosA)=cot A​

Answer 1

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

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$\sf: \implies\: {\bold{ \dfrac{ 1 + cos A - Sin^2A}{ sin A( 1 + cosA)} = CotA}}$

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$\sf: \implies\: {\bold{ \dfrac{ 1 - Sin^2A - cosA}{ sin A( 1 + cosA)} = CotA}}$

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$\sf{\red{\boxed{\bold{ 1 - Sin^2A = Cos^2A}}}}$

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$\sf: \implies\: {\bold{ \dfrac{ cos^2A - cos A }{ sin A( 1 + cosA)} = Cot A }}$

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$\sf: \implies\: {\bold{ \dfrac{ CosA ( 1 - cos A )}{ sin A( 1 + cosA)} = CotA}}$

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$\sf: \implies\: {\bold{ \dfrac{ CosA \cancel{( 1 - cos A )}}{ sin A \cancel{( 1 + cosA)}} = Cot A }}$

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$\sf: \implies\: {\bold{ \dfrac{ CosA }{ sin A} = CotA }}$

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$\sf{\red{\boxed{\bold{CotA =\dfrac{cosA}{SinA}}}}}$

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$\sf: \implies\: {\bold{ Cot A = Cot A }}$

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⠀⠀⠀⠀⠀⠀⠀⠀ .......Hence Proved

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Answer 2

$\sf{\bold{\purple{\star{\underline{\underline{Solution}}}}}}$

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$\sf : \implies\:{\frak{ \dfrac{1 + cos A - sin^2A}{sinA ( 1 + cosA)} = cot A }}$

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$\sf : \implies\:{\frak{\dfrac { 1 + sin^2A + cosA}{sinA ( 1 + cosA)} = cot A }}$

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$\sf{\star{\boxed{\orange{\frak{ 1 + sin^2A = Cos^2A}}}}}$

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$\sf : \implies\:{ \frak{\dfrac{ cos^2A - cosA}{sinA ( 1 + cosA)} = cot A }}$

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$\sf : \implies\:{\frak{ \dfrac{ Cos A ( 1 + cosA)}{sinA ( 1 + cosA)} = cot A }}$

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$\sf : \implies\:{\frak{ \dfrac{cosA}{sinA } = cot A }}$

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$\sf{\star{\boxed{\orange{\frak{ \dfrac{cosA}{SinA} = Cot A}}}}}$

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$\sf : \implies\:{\frak{ cotA = cot A }}$

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