Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$
Question.
Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$
Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$
solution:
L.H.S. $=\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}$
$=\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}$
$=\frac{-\cos ^{2} x}{-\sin ^{2} x}$
$=\cot ^{2} x$
$=$ R.HS.
L.H.S. $=\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}$
$=\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}$
$=\frac{-\cos ^{2} x}{-\sin ^{2} x}$
$=\cot ^{2} x$
$=$ R.HS.
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