Question

Solve :

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$\frac{2\sin x}{\sin 3x}+\frac{\tan x)}{\tan 3x}$

$=\frac{2\sin x}{3\sin x-4{\sin }^{3}x}+\frac{\tan x}{\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}}$

$=\frac{2\sin x}{\sin x(3-4{\sin }^{2}x)}+\frac{\tan x(1-3{\tan }^{2}x)}{\tan x(3-{\tan }^{2}x)}$

$=\frac{2}{3-4{\sin }^{2}x}+\frac{1-\frac{3{\sin }^{2}x}{{\cos }^{2}x}}{3-\frac{{\sin }^{2}x}{{\cos }^{2}x}}$

$=\frac{2}{3-4{\sin }^{2}x}+\frac{{\cos }^{2}x-3{\sin }^{2}x}{3{\cos }^{2}x-{\sin }^{2}x}$

$=\frac{2}{3-4{\sin }^{2}x}+\frac{1-{\sin }^{2}x-3{\sin }^{2}x}{3-3{\sin }^{2}x-{\sin }^{2}x}$

$=\frac{2}{3-4{\sin }^{2}x}+\frac{1-4{\sin }^{2}x}{3-4{\sin }^{2}x}$

$=\frac{2+1-4{\sin }^{2}x}{3-4{\sin }^{2}x}$

$=\frac{3-4{\sin }^{2}x}{3-4{\sin }^{2}x}=1$