Question

Let $0<x<\frac{\pi }{4}$, then $\sec 2x-\tan 2x$ is equal to

• A. $\tan (x-\frac{\pi }{4})$
• B. $\tan (\frac{\pi }{4}-x)$
• C. $\tan (x+\frac{\pi }{4})$
• D. ${\tan }^2(x+\frac{\pi }{4})$

Answer 1

The correct option is B $\tan (\frac{\pi }{4}-x)$

The value of $\sec 2x-\tan 2x$ is

$=\frac{1}{\cos 2x}-\frac{\sin 2x}{\cos 2x}$

$=\frac{1-\sin 2x}{\cos 2x}$

$=\frac{{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x}{{\cos }^{2}x-{\sin }^{2}x}$
$=\frac{{(\cos x-\sin x)}^2}{({\cos }^{2}x-{\sin }^{2}x)}=\frac{\cos x-\sin x}{\cos x+\sin x}$
$=\frac{1-\tan x}{1+\tan x}=\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}=\tan (\frac{\pi }{4}-x)$