Question:
The value of $\frac{\cot x-\tan x}{\cot 2 x}$ is__________
Solution:
$\frac{\cot x-\tan x}{\cot 2 x}$
$=\frac{\frac{1}{\tan x}-\tan x}{\frac{1}{\tan 2 x}}$
$=\frac{\frac{1-\tan ^{2} x}{\tan x}}{\frac{1}{\tan 2 x}}$
$=\frac{1-\tan ^{2} x}{\tan x} \times \tan ^{2} x$
$=\frac{1-\tan ^{2} x}{\tan x} \times \frac{2 \tan x}{1-\tan ^{2} x}$
Therefore,
$\frac{\cot x-\tan x}{\cot 2 x}=2$
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