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The function $y=\tan x-x $ is

\[\](1)an increasing function in $( 0 ,\frac{\pi}{2})$\[\](2)an decreasing function in $(0 ,\frac{\pi}{2})$\[\](3)increasing  in $(0 ,\frac{\pi}{4})$decreasing  in $(\frac{\pi}{4} ,\frac{\pi}{2})$\[\](4)decreasing  in $(0 ,\frac{\pi}{4})$ increasing in $(\frac{\pi}{4} ,\frac{\pi}{2}$)

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$y= \tan x-x$
Let $f(x) =\tan x -x$
$f'(x) =\sec^2 x -1$
$\sec \large\frac{\pi}{6}=\frac{2}{\sqrt 3}$
$\sec \large\frac{\pi}{3}$$=2$
$\sec \large\frac{\pi}{4}$$=\sqrt 2$
Hence f(x) is a increasing suction in $ (0, \large\frac{\pi}{2})$
Hence 1 is the correct answer.
answered May 20, 2014 by meena.p
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