Question

$\left[\cos ecx-\sin x\right]\left[\sec x-\cos x\right]\left[\tan x+\cot x\right]=1$

Answer 1

$=[\frac{1}{\sin x}-\sin x][\frac{1}{\cos x}-\cos x][\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}]$

$=\left[\frac{1-{\sin }^{2}x}{\sin x}\right]\left[\frac{1-{\cos }^{2}x}{\cos x}\right]\left[\frac{{\sin }^{2}x+{\cos }^{2}x}{\cos x\sin x}\right]$

$=\frac{{\cos }^{2}x}{\sin x}.\frac{{\sin }^{2}x}{\cos x}.\frac{1}{\cos x\sin x}$

$=1$