Interval in which the function $f$ given by $f(x) = \tan x - 4x, x ∈ (0,\frac{\pi}{2})$ is strictly decreasing is

$0<x<\frac{\pi}{3}$

The correct answer is Option (1) → $0<x<\frac{\pi}{3}$

$f(x) = \tan x - 4x$

$f'(x) = \sec^2 x - 4$

$f'(x) < 0 \Rightarrow \sec^2 x - 4 < 0 \Rightarrow \sec^2 x < 4 \Rightarrow \cos^2 x > \frac{1}{4} \Rightarrow \cos x > \frac{1}{2}$

$\cos x > \frac{1}{2} \Rightarrow x \in \left(0,\ \frac{\pi}{3}\right)$

$\Rightarrow f(x) \text{ is strictly decreasing in } \left(0,\ \frac{\pi}{3}\right)$