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### Answers (1)

Answer:

$f(x) \text { is increasing function on } (-\frac{\pi }{2},\frac{\pi }{2})$

Given:

$f(x) =tan \: x$

To prove:

$\text { We have to show that } f(x) \text { is increasing function on } (-\frac{\pi }{2},\frac{\pi }{2})$

Hint:

Condition for increasing function f’(x)>0.

Solution:

Given

$f(x) =tan \: x$

On differentiating both sides w.r.t x we get

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\tan x)\\ &\Rightarrow f^{\prime}(x)=\sec ^{2} x\\ \end{aligned}

Now, as given

\begin{aligned} &x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{aligned}

i.e, 4th quadrant to 1st quadrant, where

\begin{aligned} \Rightarrow sec^{2}\: x> 0 \\ \Rightarrow f'(x)> 0 \end{aligned}

Hence condition of f(x) to be increasing.

$\text { Hence } f(x) \text { is increasing on interval } x \in (-\frac{\pi }{2},\frac{\pi }{2}).$

Hence proved.

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