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If $\sin \mathrm{x}=\frac{\sqrt{5}}{3}$ and $0<\mathrm{x}<\frac{\pi}{2}$ find the values of $\tan 2 x$



To find: tan2x

From part (i) and (ii), we have

$\sin 2 x=\frac{4 \sqrt{5}}{9}$

And $\cos 2 \mathrm{x}=-\frac{1}{9}$

We know that,

$\tan x=\frac{\sin x}{\cos x}$

Replacing x by 2x, we get

$\tan 2 x=\frac{\sin 2 x}{\cos 2 x}$

Putting the values of sin 2x and cos 2x, we get

$\tan 2 x=\frac{\frac{4 \sqrt{5}}{9}}{-\frac{1}{9}}$

$\tan 2 \mathrm{x}=\frac{4 \sqrt{5}}{9} \times(-9)$

$\therefore \tan 2 x=-4 \sqrt{5}$



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