Question

$\frac{\sin ({90}^{\circ }-\theta )\cos ({90}^{\circ }-\theta )}{\cot ({90}^{\circ }-\theta )}+{\sin }^2\theta =$___

Answer 1

We know that:

$\sin ({90}^{\circ }–\theta )=\cos \theta$,

$\cos ({90}^{\circ }–\theta )=\sin \theta$ and

$\cot ({90}^{\circ }–\theta )=\tan \theta$

$\Rightarrow \frac{\cos \theta \sin \theta }{\tan \theta }+{\sin }^2\theta ={\cos }^2\theta +{\sin }^2\theta =1$