Math, asked by rishikesh7892, 10 months ago

cosA\1-tanA+Sina÷1-cotA=sinA+cosA​

Answers

Answered by harshjoisher312
0

Step-by-step explanation:

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 sandy1816
0

Step-by-step explanation:

 \frac{cosA}{1 - tanA}  +  \frac{sinA}{1 - cotA}  \\  \\  =  \frac{ {cos}^{2}A }{cosA - sinA}  +  \frac{ {sin}^{2}A }{sinA - cosA}  \\  \\  =  \frac{ {cos}^{2} A}{cosA - sinA}  -  \frac{ {sin}^{2}A }{cosA- sinA}  \\  \\  =  \frac{ {cos}^{2}A -  {sin}^{2}  A}{cosA - sinA}  \\  \\  = cosA + sinA

Similar questions