Question

1 + cos theta minus sin square theta divided by sin theta bracket 1 + cos theta bracket close equal to cot theta

Answer 1

Here $LHS=\frac{1+\cos \theta +\sin^2 \theta}{\sin \theta (1+cos \theta)}$

Now,

$LHS=\frac{1+\cos \theta -\sin^2 \theta}{\sin \theta (1+cos \theta)} \\ LHS=\frac{1+\cos \theta -(1-\cos^2 \theta)}{\sin \theta (1+cos \theta)} \\ LHS=\frac{1+\cos \theta -(1+\cos \theta )(1-\cos \theta )}{\sin \theta (1+cos \theta)} \\ LHS=\frac{1 -(1-\cos \theta )}{\sin \theta } \\ LHS=\frac{\cos \theta }{\sin \theta } \\ LHS=\cot \theta = RHS$

The proof is complete.

Answer 2

Answer:

Solution is given at the below in the image

Step-by-step is given kn it