Question

Let $0<x<\frac{\pi }{4}$, then the value of $(\sec 2x-\tan 2x)$ is

• 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 $\tan (\frac{\pi }{4}-x)$

$\sec 2x-\tan 2x=\frac{1-\sin 2x}{\cos 2x}=\frac{1-2\sin x\cos x}{{\cos }^{2}x-{\sin }^{2}x}=\frac{(\cos x-\sin x{)}^2}{(\cos x-\sin x)(\cos x+\sin x)}=\frac{\cos x-\sin x}{\cos x+\sin x}(\because \cos x\ne \sin x\Rightarrow \cos x-\sin x\ne 0)=\frac{1-\tan x}{1+\tan x}(\because \cos x\ne 0)$
$=\tan (\frac{\pi }{4}-x)$