Math, asked by cbanu2005, 10 months ago

cosA÷1-tanA+sinA÷1-cotA=sin A+cos A​

Answers

Answered by alok18092
1

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 harpreetsingh4866
1

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

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