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Prove that,
$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$
$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$
Solution:
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Verified Answer
$\mathrm{LHS}=\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)$
$\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$
Now, $\cos \left(\frac{3 \pi}{2}+x\right)=\sin x, \cos (2 \pi+x)=\cos x$ and $\cot \left(\frac{3 \pi}{2}-x\right)=\tan x, \cot (2 \pi+x)=\cot x$
L.H.S. $=\sin x \cdot \cos x[\tan x+\cot x]$
$=\sin x \cdot \cos x\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]$
$=\sin x \cdot \cos x\left[\frac{\sin ^2 x+\cos ^2 x}{\cos x \sin x}\right]$
$=(\sin x \cdot \cos x) \frac{1}{\cos x \sin x}=1$
$\left[\because \sin ^2 x+\cos ^2 x=1\right]$
$\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$
Now, $\cos \left(\frac{3 \pi}{2}+x\right)=\sin x, \cos (2 \pi+x)=\cos x$ and $\cot \left(\frac{3 \pi}{2}-x\right)=\tan x, \cot (2 \pi+x)=\cot x$
L.H.S. $=\sin x \cdot \cos x[\tan x+\cot x]$
$=\sin x \cdot \cos x\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]$
$=\sin x \cdot \cos x\left[\frac{\sin ^2 x+\cos ^2 x}{\cos x \sin x}\right]$
$=(\sin x \cdot \cos x) \frac{1}{\cos x \sin x}=1$
$\left[\because \sin ^2 x+\cos ^2 x=1\right]$
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