[JEE Advanced 1994] for 0 x pi 4 sec 2x tan 2x equals
For $0<x<\frac\pi4$, $\sec2x-\tan2x$ equals
(a) $\tan(x-\frac\pi4)$
(b) $\tan(\frac\pi4-x)$
(c) $\tan(x+\frac\pi4)$
(d) $\tan^2(x+\frac\pi4)$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
Correct answer: (b) $\tan(\frac\pi4-x)$
$\sec2x-\tan2x=\frac{1-\sin2x}{\cos2x}=\frac{(\cos x-\sin x)^2}{\cos^2x-\sin^2x}=\frac{\cos x-\sin x}{\cos x+\sin x}=\tan\!\left(\frac\pi4-x\right).$
JEE Advanced 1994 · Trigonometry — verified solution by the Vidaara Team.
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