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Questions  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Q)

If $\tan(\cot x)=\cot(\tan x)$,prove that $\sin 2x=\large\frac{4}{(2n+1)\pi}$

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A)
Toolbox:
  • $\tan(\large\frac{\pi}{2}$$-\theta)=\cot \theta$
  • $\tan \theta=\tan \alpha\Rightarrow \theta=n\pi+\alpha$
  • $\cos^2x+\sin^2x=1$
  • $2\sin A.\cos A=\sin 2A$
$\tan(\cot x)=\cot(\tan x)$
$\tan(\cot x)=\tan (\large\frac{\pi}{2}$$-\tan x)$
$\cot \theta=\tan (\large\frac{\pi}{2}$$-\theta$)
$\cot x=n\pi+(\large\frac{\pi}{2}$$-\tan x)$
$\tan \theta=\tan \alpha\Rightarrow \theta=n\pi+\alpha$
$\cot x+\tan x=n\pi+\large\frac{\pi}{2}$
$\large\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}=\large\frac{\pi}{2}$$(2n+1)$
$\large\frac{\cos^2x+\sin^2x}{\sin x\cos x}=\large\frac{\pi}{2}$$(2n+1)$
$\large\frac{1}{\sin x\cos x}$$=(2n+1)\large\frac{\pi}{2}$
$\large\frac{1}{2\sin x.\cos x}=$$(2n+1)\large\frac{\pi}{4}$
$\large\frac{1}{\sin 2x}=$$(2n+1)\large\frac{\pi}{4}$
$\sin 2x=\large\frac{4}{(2n+1)\pi}$
Hence proved
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