Question

cosA/1-tanA + SinA /1-cotA = cosA +sinA

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Answer 1

answer:

=cosA/(1-tanA)+sinA/(1-cotA)

=cos A/(1 - sin A/cos A) + sin A/(1 - cos A/sin A)

=cos²A/ (cos A - sin A) + sin²A / (sin A - cos A)

=cos²A/ (cos A - sin A) - sin²A / (cos A - sin A)

=(cos ² A - sin ² A) / (cos A - sin A)

=(cos A - sin A)(cos A + sin A) / (cos A - sin A)

=cos A + sin A i.e RHS

Answer 2

$\sf{\underline{\boxed{\purple{\large{\bold{ Solution }}}}}}$

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$\sf :\implies\:{\bold{\dfrac{CosA}{ 1 - TanA} + \dfrac{SinA}{ 1 - CotA} = CosA + SinA }}$

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

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

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$\sf :\implies\:{\bold{\dfrac{CosA}{ 1 - \dfrac{SinA}{CosA}} + \dfrac{SinA}{ 1 - \dfrac{CosA}{SinA}} = CosA + SinA }}$

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

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

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

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

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

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

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

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$\sf{\red{\boxed{\bold{a^2 + b^2 = ( a + b) ( a - b ) }}}}$

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

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

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LHS = RHS

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀......Hence Verified

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