Trigonometric Identities Question 290
Question: Let $ 0<x<\frac{\pi }{4}. $ Then $ \sec 2x-\tan 2x= $
[IIT Screening 1994]
Options:
A) $ \tan ( x-\frac{\pi }{4} ) $
B) $ \tan ( \frac{\pi }{4}-x ) $
C) $ \tan ( x+\frac{\pi }{4} ) $
D) $ {{\tan }^{2}}( x+\frac{\pi }{4} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \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 ( \frac{\pi }{4} )\sin x}=\tan ( \frac{\pi }{4}-x ) $ .
BETA
