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