Question

If $\sin x=\frac{1}{3}$ and $\cos x<0$, then the value of $\tan 3x$ is

• A. $\frac{23}{10\sqrt{2}}$
• B. $-\frac{23}{10\sqrt{2}}$
• C. $-\frac{23}{5\sqrt{2}}$
• D. $-\frac{23}{15\sqrt{2}}$

Answer 1

The correct option is B $-\frac{23}{10\sqrt{2}}$

$\sin x=\frac{1}{3}$
As $\sin x$ is positive and $\cos x$ is negative, so $x$ lies in the second quadrant.
$\Rightarrow \tan x=-\frac{1}{2\sqrt{2}}$
$\tan 3x=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}=\frac{-\frac{3}{2\sqrt{2}}+\frac{1}{16\sqrt{2}}}{1-\frac{3}{8}}=\frac{-\frac{23}{16\sqrt{2}}}{\frac{5}{8}}=-\frac{23}{10\sqrt{2}}$