Question

If ${\log }_2\sin x-{\log }_2\cos x-{\log }_2(1-{\tan }^{2}x)=-1$, then

• A. $\frac{n\pi }{2}+\frac{\pi }{8},n\in Z$
• B. $n\pi -\frac{\pi }{8},n\in Z$
• C. $\frac{n\pi }{4}+\frac{\pi }{8},n\in Z$
• D. $Noneofthese$

Answer 1

Answer: (A)

${\log }_2\sin x-{\log }_2\cos x-{\log }_2(1-{\tan }^{2}x)=-1$
$\Rightarrow$ ${\log }_2\frac{\sin x/\cos x}{1-{\tan }^{2}x}=-1$
$\Rightarrow$ ${\log }_2\left(\frac{\tan x}{1-{\tan }^{2}x}\right)=-1$
$\therefore$ $\frac{\tan x}{1-{\tan }^{2}x}=2^{-1}$
$\Rightarrow$ $\frac{2\tan x}{1-{\tan }^{2}x}=2^{-1}.2^{-1}=2^{\circ }=1$
$\Rightarrow$ $\tan 2x=1=\tan \frac{\pi }{4}$
$\therefore$ $2x=n\pi +\frac{\pi }{4}$
$\Rightarrow$ $x=\frac{n\pi }{2}+\frac{\pi }{8},n\pi Z$