Question

If $x\in (\frac{\pi }{4},\frac{\pi }{2})$ ,then $(\sec 2x-\tan 2x)$ equals

• A. $\tan (\frac{\pi }{4}+x)$
• B. $\tan (\frac{\pi }{4}-x)$
• C. $\tan (x-\frac{\pi }{4})$
• D. None of these

Answer 1

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

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

Hence, option 'B' is correct.