Question

Show that f(x) = tan x is an increasing function on (−π/2, π/2).

Answer 1

$$\begin{gathered} f\left(x\right)=\tan x \\ f\text{'}\left(x\right)={\sec }^{2}x \\ \text{Here,} \\ \frac{-\mathrm{\pi }}{2}<x<\frac{\mathrm{\pi }}{2} \\ \Rightarrow \sec x>0\left[\because \mathrm{Sec}\mathrm{function}\mathrm{is}\mathrm{positive}\mathrm{in}\mathrm{first}\mathrm{and}\mathrm{fourth}\mathrm{quadrant}\right] \\ \Rightarrow {\sec }^{2}x>0 \\ \Rightarrow f\text{'}\left(x\right)>0,\forall x\in \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right). \end{gathered}$$