Which of the following functions are strictly decreasing on $\left( {0,\cfrac{\pi }{2}} \right)$?

(A) $\cos x$

(B) $\cos 2x$

(C) $\cos 3x$

(D) $\tan x$

Let us see each option one-by-one.

(A) Let $f(x) = \cos x$, then

$f'(x) = - \sin x < 0$ for all $x \in \left( {0,\cfrac{\pi }{2}} \right)$

$\Rightarrow f$ is strictly decreasing on $\left( {0,\cfrac{\pi }{2}} \right)$

(B) Let $f(x) = \cos 2x$ , then
$f'(x) = - 2\sin 2x < 0$ for all $x \in \left( {0,\cfrac{\pi }{2}} \right)$
( in $(0,\;\pi ) \Rightarrow \sin 2x > 0$ in $(0,\pi /2)$)
$\Rightarrow f$ is strictly decreasing on $\left( {0,\cfrac{\pi }{2}} \right)$

(C) Let$f(x) = \cos 3x$ , then $f'(x) = - 3\sin 3x,$
which assumes $+ ve$ as well as $- ve$ values in $\left( {0,\cfrac{\pi }{2}} \right)$.$\left[ {{\rm{if}}0 < x < \cfrac{\pi }{2} \Rightarrow 0 < 3x < \cfrac{{3\pi }}{2} \Rightarrow \sin 3x > 0} \right.$ in $\left( {0,\cfrac{\pi }{3}} \right)$ and $\sin 3x < 0$ in

$\left. {\left( {\cfrac{\pi }{3},\cfrac{\pi }{2}} \right)} \right]$
$\therefore f$ is neither increasing nor decreasing on $\left( {0,\cfrac{\pi }{2}} \right)$

(D) Let $f(x) = \tan x$, then $f'(x) = {\sec ^2}x > 0$ for all $x \in \left( {0,\cfrac{\pi }{2}} \right)$

$\Rightarrow f$ is strictly increasing on $\left( {0,\cfrac{\pi }{2}} \right)$