Prove that
$\sin 2 x(\tan x+\cot x)=2$
To Prove: $\sin 2 x(\tan x+\cot x)=2$
Taking LHS,
sin 2x(tan x + cot x)
We know that,
$\tan \theta=\frac{\sin \theta}{\cos \theta} \& \cot \theta=\frac{\cos \theta}{\sin \theta}$
$=\sin 2 x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)$
$=\sin 2 x\left(\frac{\sin x(\sin x)+\cos x(\cos x)}{\cos x \sin x}\right)$
$=\sin 2 x\left(\frac{\sin ^{2} x+\cos ^{2} x}{\cos x \sin x}\right)$
We know that,
$\sin 2 x=2 \sin x \cos x$
$=2 \sin x \cos x\left(\frac{\sin ^{2} x+\cos ^{2} x}{\cos x \sin x}\right)$
$=2\left(\sin ^{2} x+\cos ^{2} x\right)$
$=2 \times 1\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]$
$=2$
= RHS
∴ LHS = RHS
Hence Proved
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