Question

If $\log_2 \: \sin x - \log_2 \cos x -\log_2(1 - \tan^2x) = - 1$, 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. $None\, of\, these$

Answer 1

Answer: (A)

$\frac{n\pi}{2}+\frac{\pi}{8},n\in\,Z$

Answer 2

$\log_2 \sin x - \log_2 \cos x - \log_2 \, (1 - \tan^2 \, x) = - 1$
$\Rightarrow$ $\log_2 \frac{\sin \, x/\cos\, x}{ 1 -\tan^2 \, x} = -1$
$\Rightarrow$ $\log_2 \left( \frac{ \tan \, x}{1 - \tan^2 \, x} \right) = - 1$
$\therefore$ $\frac{\tan \, x}{1 - \tan^2 \,x} = 2^{-1}$
$\Rightarrow$ $\frac{2 \, \tan \, x}{1 - \tan^2 \, x} = 2^{-1}.2^{-1} = 2^\circ = 1$
$\Rightarrow$ $\tan 2x = 1 = \tan \frac{\pi}{4}$
$\therefore$ $2 x = n \pi + \frac{\pi}{4}$
$\Rightarrow$ $x = \frac{n \pi}{2} + \frac{\pi}{8} , n \pi Z$