Question

if $\frac{\tan 3x}{\tan x}=2$ then prove that $\Rightarrow \sin x=\sqrt{\frac{8}{5}}$

Answer 1

$\frac{\tan 3x}{\tan x}=2$

$\Rightarrow \frac{3\tan \left(x\right)-{\tan }^{3}x}{\tan x(3-3{\tan }^{2}x)}=2$

$\Rightarrow \frac{\tan x(3-{\tan }^{2}x)}{\tan x.3(1-{\tan }^{2}x)}=2$

$\Rightarrow \frac{(2+1-{\tan }^{2}x)}{3(1-{\tan }^{2}x)}=2$

$\Rightarrow \frac{2}{3(1-{\tan }^{2}x)}+\frac{(1-{\tan }^{2}x)}{3(1-{\tan }^{2}x)}=2$

$\Rightarrow \frac{2}{3(1-{\sec }^{2}x+1)}+\frac{1}{3}=2$

$\Rightarrow \frac{2}{3(2-{\sec }^{2}x)}=2-\frac{1}{3}$

$\Rightarrow \frac{2}{3(2-{\sec }^{2}x)}=\frac{5}{3}$

$\Rightarrow -5{\sec }^{2}x=-8$

$\Rightarrow {\sec }^{2}x=\frac{-8}{5}$

$\Rightarrow \sin x=\sqrt{\frac{8}{5}}$

Hence, the answer is $\sqrt{\frac{8}{5}}.$