Math, asked by beharatanusri, 11 months ago

prove that cosA/1-tanA+sinA/1-cotA=sinA+cosA

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Answered by eena6992

Answer:

LHS

=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

Answered by chaitragouda8296

To Prove :

$\frac{cos}{1 - tan} + \frac{sin}{1 - cot} = sin \: + cos$

Solution :

LHS =

$= \frac{cos}{1 - tan} + \frac{sin}{1 - cot} \\ \\ = \frac{cos}{1 - \frac{sin}{cos} } + \frac{sin}{1 - \frac{cos}{sin} } \\ \\ = \frac{cos}{ \frac{cos - sin}{cos} } + \frac{sin}{ \frac{sin - cos}{sin} } \\ \\ = cos \times \frac{cos}{cos - sin} + sin \times \frac{sin}{sin - cos} \\ \\ = \frac{ {cos}^{2} }{cos - sin} + \frac{ {sin}^{2} }{sin - cos} \\ \\ = \frac{ {cos}^{2} }{cos - sin} + \frac{ {sin}^{2} }{ - ( - sin + cos) }\\ \\ = \frac{ {cos}^{2} }{(cos - sin) } - \frac{ {sin}^{2} }{(cos - sin) } \\ \\ = \frac{ {cos}^{2} - {sin }^{2} }{cos - sin} \\ \\ Here , numerator \: \: \: is \: \: \: in \: \: \: the \: \: \: form \: \: \:of \: \: \: \\ {x}^{2} + {y}^{2} = (x + y)(x - y) \\ \\ = \frac{(cos \: + \: sin)(cos - sin)}{cos - sin} \\ \\ Here, (cos \: - sin \: ) \: gets \: \: \: cancelled \\ \\ = cos \: + sin \\ \\ = sin \: + \: cos \\ \\$

= RHS

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prove that cosA/1-tanA+sinA/1-cotA=sinA+cosA​

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To Prove :

Solution :

prove that cosA/1-tanA+sinA/1-cotA=sinA+cosA