Question

If ${\log }_2(3\sin x)-{\log }_2(\cos x)-{\log }_2(1-\tan x)-{\log }_2(1+\tan x)=1$, then $\tan x$ is equal to

• A. -2
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

We have
${\log }_2(3\sin x)-{\log }_2(\cos x)-{\log }_2(1-\tan x)-{\log }_2(1+\tan x)=1$
$\Rightarrow {\log }_2\left(\frac{3\sin x}{\cos x(1-\tan x)(1+\tan x)}\right)=1\Rightarrow {\log }_2\left(\frac{3\tan x}{1-{\tan }^{2}x}\right)=1$
$\Rightarrow 3\tan x=2\left(1-{\tan }^{2}x\right)\Rightarrow 2{\tan }^{2}x+3\tan x-2=0$
$\Rightarrow 2{\tan }^{2}x+4\tan x-\tan x-2=0(2\tan x-1)(\tan x+2)=0$
$\Rightarrow \tan x=\frac{1}{2}\text{ or }\tan x=-2(\text{ rejected, think!) }$