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Prove : $\large\frac{\cos(\pi+x)\cos(-x)}{\sin (\pi-x)\cos(\large\frac{\pi}{2}+\normalsize x)}=$$\cot^2x$

1 Answer

Toolbox:
  • $\cos(\pi+x)=-\cos x,\cos(-x)=\cos x$
  • $\sin (\pi-x)=\sin x,\cos(\large\frac{\pi}{2}+$$x)=-\sin x$
L.H.S
$\large\frac{\cos(\pi+x)\cos(-x)}{\sin (\pi-x)\cos(\large\frac{\pi}{2}+\normalsize x)}=\frac{-\cos x.\cos x}{\sin x.-\sin x}$
$\Rightarrow \large\frac{-\cos^2x}{-\sin^2x}$
$\Rightarrow \cot^2x$=R.H.S
Hence proved
answered Apr 17, 2014 by sreemathi.v
 

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