Question

Show that function \( f(x) = \tan x \) is increasing in \( (0, \frac{\pi}{2}) \).

Hint:

A function is increasing if \( f'(x) > 0 \); use trigonometric identities to confirm derivative's sign.

Answer 1

A function is increasing if its derivative \( f'(x) > 0 \) in the interval.
\[ f(x) = \tan x, f'(x) = \sec^2 x. \] Since \( \sec^2 x = \frac{1}{\cos^2 x} \), and \( \cos x > 0 \) in \( (0, \frac{\pi}{2}) \),
\[ \sec^2 x > 0 \text{ for all } x \in (0, \frac{\pi}{2}). \] Thus, \( f'(x) > 0 \), so \( f(x) = \tan x \) is increasing in \( (0, \frac{\pi}{2}) \).
Answer: \( \tan x \) is increasing since \( \sec^2 x > 0 \).