Answer:
Option D
Explanation:
$\lim_{x \rightarrow {\pi/4}}\frac{\tan x-1}{\cos 2x}=\lim_{h \rightarrow 0}\frac{\tan(\frac{\pi}{4}+h)-1}{\cos2(\frac{\pi}{4}+h)}$
[$\because$ $x=\frac{\pi}{4}+h$]
=$\lim_{h \rightarrow 0}\frac{(\frac{1+\tan h}{1-\tan h})-1}{cos (\frac{\pi}{2}+2h)}$
=$\lim_{h \rightarrow 0}\frac{1+\tanh-1+\tan h}{-\sin 2h(1-\tan h)}$
= $\lim_{h \rightarrow 0}\frac{-2\tanh}{2\sin h\cos h(1-\tan h)}$
= $\lim_{h \rightarrow 0}\frac{-1}{\cos^{2}h(1-\tan h)}=-1$
