Question

Find the value of $\frac{\sin \theta }{\cos ({90}^0+\theta )}+\frac{\sin \theta }{\sin ({180}^0+\theta )}+\frac{\tan ({90}^0+\theta )}{\cot \theta }$.

Answer 1

The given expression is equal to
$\frac{\sin \theta }{-\sin \theta }+\frac{\sin \theta }{-\sin \theta }+\frac{-\cot \theta }{\cot \theta }$
$=-1-1-1=-3$