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Q. Assertion $: \frac{\cos (\pi+x) \cdot \cos (-x)}{\sin (\pi-x) \cdot \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$
Reason $: \cos (\pi+\theta)=-\cos\, \theta$ and $\cos (-\theta)=\cos \,\theta$
Also, $\sin (\pi-\theta)=\sin \,\theta$ and $\sin (-\theta)=-\sin \,\theta$

Trigonometric Functions

Solution:

$\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}$
$=\frac{\cos ^{2} x}{\sin ^{2} x}=\cot ^{2} x$