If ${log}_{_{2}} (sin x) - {log}_{2} (cos x) - {log}_{2} (1 - tan x) - {log}_{2} (1 + tan x) = - 1$ , then $t a n 2 x =$

- 1

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Solution

The correct option is B $1$
$l o g_{2} (sin x) - {log}_{2} (cos x) - l o g_{2} (1 - tan x) - l o g_{2} (1 + tan x) = - 1$

$\Rightarrow {log}_{2} (\frac{sin x}{cos x (1 - tan x) (1 + tan x)}) = - 1$

$\Rightarrow {log}_{2} (\frac{tan x}{1 - {tan}^{2} x}) = - 1$

$\Rightarrow \frac{tan x}{1 - {tan}^{2} x} = 2^{- 1} = \frac{1}{2}$

$\Rightarrow \frac{2 tan x}{1 - {tan}^{2} x} = 2 \times \frac{1}{2} = 1$

$tan 2 x = 1$

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