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  4. If log2 : sin x - log2 cos x - log2(1 - tan2x) = - 1, then

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Q. If $\log_2 \: \sin x - \log_2 \cos x -\log_2(1 - \tan^2x) = - 1$, then

COMEDKCOMEDK 2008Trigonometric Functions

Solution:

$\log_2 \sin x - \log_2 \cos x - \log_2 \, (1 - \tan^2 \, x) = - 1$
$\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 $ $x = \frac{n \pi}{2} + \frac{\pi}{8} , n \pi Z$