[tex]csc(\theta)=\cfrac{1}{sin(\theta)}~\hspace{10em} sin(2\theta)=2sin(\theta)cos(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{csc^2(x)}{2cot(x)}=cot(2x) \\\\[-0.35em] ~\dotfill\\\\ \cfrac{csc^2(x)}{2cot(x)}\implies \cfrac{~~\frac{1}{sin^2(x)} ~~}{2\cdot \frac{cos(x)}{sin(x)}}\implies \cfrac{1}{sin^2(x)}\cdot \cfrac{sin(x)}{2cos(x)}\implies \cfrac{1}{2sin(x)cos(x)} \\\\\\ \cfrac{1}{sin(2x)}\implies csc(2x)[/tex]
Step-by-step explanation:
Recall that
[tex]\csc{x} = \dfrac{1}{\sin{x}}[/tex]
[tex]\cot{x} = \dfrac{1}{\tan{x}} = \dfrac{\cos{x}}{\sin{x}}[/tex]
so we can write the original expression as
[tex]\dfrac{\csc^2x}{2\cot{x}} = \dfrac{1}{\sin^2x}\cdot\dfrac{1}{2\cot{x}} = \dfrac{1}{\sin^2x}\cdot\dfrac{\sin{x}}{2\cos{x}}[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:=\dfrac{1}{2\sin{x}\cos{x}}[/tex] (1)
But recall also the identity
[tex]2\sin{x}\cos{x} = \sin2{x}[/tex]
so we can rewrite Eqn(1) as
[tex]\dfrac{\csc^2x}{2\cot{x}} = \dfrac{1}{\sin2{x}} = \csc2{x}[/tex]
