Question
The function $\text{f(x)}=\tan\text{x}-\text{x}$
Solution:
We have, $\text{f(x)}=\tan\text{x}-\text{x}$
$\therefore\ \text{f}'(\text{x})=\sec^2\text{x}-1$
Since, $\text{f}'(\text{x})>0,\forall\text{ x}\in\text{R}$
Hence, f(x) always increases.
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