MCQ
If $0 < x<\frac{\pi}{4}$, then $\sec 2 x-\tan 2 x=$
  • A
    $\tan \left(x-\frac{\pi}{4}\right)$
  • $\tan \left(\frac{\pi}{4}-x\right)$
  • C
    $\tan \left(x+\frac{\pi}{4}\right)$
  • D
    $\tan ^2\left(x+\frac{\pi}{4}\right)$

Answer

Correct option: B.
$\tan \left(\frac{\pi}{4}-x\right)$
(B)
$\sec 2 x-\tan 2 x=\frac{1-\sin 2 x}{\cos 2 x}=\frac{(\cos x-\sin x)^2}{\left(\cos ^2 x-\sin ^2 x\right)}$
$=\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{1-\tan x}{1+\tan x}$
$=\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \binom{\pi}{4} \tan x}=\tan \left(\frac{\pi}{4}-x\right)$

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