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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions

$\cos (\large\frac{3\pi}{2}+$$x)\cos(2\pi+x)[\cot(\large\frac{3\pi}{2}$$-x)+\cot(2\pi+x)]=1$

1 Answer

Toolbox:
  • $\cos(\large\frac{3\pi}{2}$$+x)=\sin x$
  • $\cos (2\pi+x)=\cos x$
  • $\cot(\large\frac{3\pi}{2}$$-x)=\tan x$
  • $\cot (2\pi+x)=\cot x$
L.H.S
$\cos (\large\frac{3\pi}{2}+$$x)\cos(2\pi+x)[\cot(\large\frac{3\pi}{2}$$-x)+\cot(2\pi+x)]$
$\Rightarrow \sin x.\cos x[\tan x+\cot x]$
$\Rightarrow \sin x\cos x[\large\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}]$
$\Rightarrow \sin x\cos x[\large\frac{\sin^2 x+\cos ^2 x}{\sin x\cos x}]$
$\Rightarrow \sin^2x+\cos^2x=1$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 

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