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Question: Let\(0 < x < \frac{\pi}{4}.\) Then \(\sec 2x - \tan 2x =\)...

Let0<x<π4.0 < x < \frac{\pi}{4}. Then sec⁡2x−tan⁡2x=\sec 2x - \tan 2x =

A

tan⁡(x−π4)\tan\left( x - \frac{\pi}{4} \right)

B

tan⁡(π4−x)\tan\left( \frac{\pi}{4} - x \right)

C

tan⁡(x+π4)\tan\left( x + \frac{\pi}{4} \right)

D

tan⁡2(x+π4)\tan^{2}\left( x + \frac{\pi}{4} \right)

Answer

tan⁡(π4−x)\tan\left( \frac{\pi}{4} - x \right)

Explanation

Solution

sec⁡2x−tan⁡2x=1−sin⁡2xcos⁡2x\sec 2x - \tan 2x = \frac{1 - \sin 2x}{\cos 2x}

=(cos⁡x−sin⁡x)2(cos⁡2x−sin⁡2x)=cos⁡x−sin⁡xcos⁡x+sin⁡x=1−tan⁡x1+tan⁡x= \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}

=tan⁡π4−tan⁡x1+tan⁡(π4)sin⁡x=tan⁡(π4−x)= \frac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\left( \frac{\pi}{4} \right)\sin x} = \tan\left( \frac{\pi}{4} - x \right).