If $$f\left( x \right) = \frac{x}{{\sin x}}$$    and $$g\left( x \right) = \frac{x}{{\tan x}}$$    where $$0 < x \leqslant 1,$$   then in this interval

Solution :
$$\eqalign{ & {\text{We}}\,{\text{have}}\,f\left( x \right) = \frac{x}{{\sin x}},0 < x \leqslant 1 \cr & \Rightarrow f'\left( x \right) = \frac{{\sin x - x\cos x}}{{{{\sin }^2}x}} \cr & {\text{where}}\,{\sin ^2}x{\text{ is always }} + ve,{\text{ when }}0 < x \leqslant 1.{\text{ But to check Nr}}{\text{., we again let}} \cr & h\left( x \right) = \sin x - x\cos x \cr & \Rightarrow h'\left( x \right) = x\sin x > 0\,{\text{for}}\,0 < x \leqslant 1 \Rightarrow h\left( x \right)\,{\text{is increasing}} \cr & \Rightarrow h\left( 0 \right) < h\left( x \right),\,{\text{when}}\,0 < x \leqslant 1 \cr & \Rightarrow 0 < \sin x - x\cos x,\,{\text{when}}\,0 < x \leqslant 1 \cr & \Rightarrow \sin x - x\cos x > 0,\,{\text{when}}\,0 < x \leqslant 1 \cr & \Rightarrow f'\left( x \right) > 0,x \in \left( {0,1} \right] \cr & \Rightarrow f\left( x \right)\,{\text{is}}\,{\text{increasing}}\,{\text{on}}\,\left( {0,1} \right] \cr & {\text{Again}}\,g\left( x \right) = \frac{x}{{\tan x}} \cr & \Rightarrow g'\left( x \right) = \frac{{\tan x - x{{\sec }^2}x}}{{{{\tan }^2}x}},\,{\text{when}}\,0 < x \leqslant 1 \cr & {\text{Here}}\,{\tan ^2}x > 0\,{\text{But}}\,{\text{to check Nr}}{\text{. we consider}} \cr & p\left( x \right) = \tan x - x{\sec ^2}x \cr & p'\left( x \right) = {\sec ^2}x - {\sec ^2}x - x.2\sec x.\sec x\tan x \cr & \Rightarrow p'\left( x \right) = - 2x{\sec ^2}x\tan x < 0\,{\text{for}}\,0 < x \leqslant 1 \cr & \Rightarrow p\left( x \right)\,{\text{is decreasing, when}}\,0 < x \leqslant 1 \cr & \Rightarrow p\left( 0 \right) > p\left( x \right) \Rightarrow 0 > \tan x - x{\sec ^2}x \cr & \therefore g'\left( x \right) < 0 \cr & {\text{Hence }}g\left( x \right){\text{ is decreasing when}}\,0 < x \leqslant 1. \cr} $$