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3.Trigonometrical Ratios, Functions and Identities
easy
Let $0 < x < \frac{\pi }{4}.$ Then $\sec 2x - \tan 2x = $
A
$\tan \left( {x - \frac{\pi }{4}} \right)$
B
$\tan \left( {\frac{\pi }{4} - x} \right)$
C
$\tan \left( {x + \frac{\pi }{4}} \right)$
D
${\tan ^2}\left( {x + \frac{\pi }{4}} \right)$
(IIT-1994)
Solution
(b) $\sec 2x – \tan 2x = \frac{{1 – \sin 2x}}{{\cos 2x}}$
$ = \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)$.
Standard 11
Mathematics
