Question

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

• A. $\tan \theta$
• B. 1
• C. -1
• D. $\sin \theta$

Answer 1

The correct option is B 1

We have,

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

$\Rightarrow \frac{\cos \theta \sin \theta }{\frac{\sin \theta }{\cos \theta }}+\frac{\sin \theta \cos \theta }{\frac{\cos \theta }{\sin \theta }}$

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

$\Rightarrow {\cos }^2\theta +{\sin }^2\theta$

$\Rightarrow 1$

Hence, this is the answer.