Question

Prove that:
CosA/ 1 - tanA + sin^2 A/ sinA - cos A = sinA + cosA

Answer 1

cos(A)1−tan(A)+sin(A)1−cot(A)cos⁡(A)1−tan⁡(A)+sin⁡(A)1−cot⁡(A)

Express in terms of sinsin and coscos

=cos(A)1−sin(A)cos(A)+sin(A)1−cos(A)sin(A)=cos⁡(A)1−sin⁡(A)cos⁡(A)+sin⁡(A)1−cos⁡(A)sin⁡(A)

Let us now simplify cos(A)1−sin(A)cos(A)cos⁡(A)1−sin⁡(A)cos⁡(A)

For the denominator 1−sin(A)cos(A)1−sin⁡(A)cos⁡(A), we convert this to =cos(A)−sin(A)cos(A)=cos⁡(A)−sin⁡(A)cos⁡(A)

resulting to =cos(A)cos(A)−sin(A)cos(A)=cos⁡(A)cos⁡(A)−sin⁡(A)cos⁡(A)

Applying the fraction rule abc=a⋅cbabc=a⋅cb

We have =cos(A)cos(A)cos(A)−sin(A)→=cos2(A)cos(A)−sin(A)=cos⁡(A)cos⁡(A)cos⁡(A)−sin⁡(A)→=cos2⁡(A)cos⁡(A)−sin⁡(A)

For sin(A)1−cos(A)sin(A)sin⁡(A)1−cos⁡(A)sin⁡(A), we use the same processes as above, resulting in

=sin(A)sin(A)−cos(A)sin(A)→=sin(A)sin(A)sin(A)−cos(A)→=sin2(A)sin(A)−cos(A)=sin⁡(A)sin⁡(A)−cos⁡(A)sin⁡(A)→=sin⁡(A)sin⁡(A)sin⁡(A)−cos⁡(A)→=sin2⁡(A)sin⁡(A)−cos⁡(A)

So now we go back to this equation: =cos2(A)cos(A)−sin(A)+sin2(A)sin(A)−cos(A)=cos2⁡(A)cos⁡(A)−sin⁡(A)+sin2⁡(A)sin⁡(A)−cos⁡(A)

Using the LHS cos(A)−sin(A)cos⁡(A)−sin⁡(A), we have this equation:

=cos2(A)cos(A)−sin(A)−sin2(A)cos(A)−sin(A)→cos2(A)−sin2(A)cos(A)−sin(A)
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