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$\lim\limits_{x\to\large\frac{-\pi}{4}}\large\frac{1+\tan x}{\cos 2x}$=

$(a)\;1\qquad(b)\;0\qquad(c)\;-2\qquad(d)\;-1$

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$\lim\limits_{x\to -\large\frac{-\pi}{4}}\large\frac{1+\tan x}{\cos 2x}\big(\frac{0}{0}\big)$
$\Rightarrow \lim\limits_{x\to \large\frac{-\pi}{4}}\large\frac{\sec^2x}{-2\sin 2x}$
$\Rightarrow \large\frac{\sec^2(-\Large\frac{\pi}{4}\big)}{-2\sin(-\Large\frac{\pi}{2})}$
$\Rightarrow 1$
Hence (a) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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