MCQ
જો $0 < x < \frac{\pi }{4}.$ તો $\sec 2x - \tan 2x = $
- A$\tan \left( {x - \frac{\pi }{4}} \right)$
- ✓$\tan \left( {\frac{\pi }{4} - x} \right)$
- C$\tan \left( {x + \frac{\pi }{4}} \right)$
- D${\tan ^2}\left( {x + \frac{\pi }{4}} \right)$
$ = \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 \left( {\frac{\pi }{4}} \right)\sin x}} = \tan \left( {\frac{\pi }{4} - x} \right)$.
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