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) \[{{\tan }^{2}}\left( x+\frac{\pi }{4} \right)\]
Correct Answer: B
Solution :
\[\sec 2x-\tan 2x=\frac{1-\sin 2x}{\cos 2x}\] \[=\frac{{{(\cos x-\sin x)}^{2}}}{({{\cos }^{2}}x-{{\sin }^{2}}x)}=\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{1-\tan x}{1+\tan x}\] \[=\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \left( \frac{\pi }{4} \right)\sin x}=\tan \left( \frac{\pi }{4}-x \right)\].
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