Answer:
Trig identities
[tex]\tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)}[/tex]
[tex]\sin(\thet2\theta)=2 \sin(\theta)\cos(\theta)[/tex]
[tex]\begin{aligned}\cos(\thet2\theta) & =1-2\sin^2(\theta)\\2\sin^2(\theta) & = 1-\cos(2\theta) \end{aligned}[/tex]
Solution
[tex]\tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)}[/tex]
[tex]\textsf{Multiply by}\:\dfrac{2\sin(\theta)}{2\sin(\theta)}:[/tex]
[tex]\begin{aligned}\implies \tan(\theta) & =\dfrac{\sin(\theta)}{\cos(\theta)} \times \dfrac{2\sin(\theta)}{2\sin(\theta)}\\\\ & =\dfrac{2\sin^2(\theta)}{2\sin(\theta)\cos(\theta)}\\\\ & = \dfrac{1-\cos(2\theta)}{\sin(2\theta)}\end{aligned}[/tex]
Therefore,
[tex]\implies \dfrac{1-\cos(2\theta)}{\sin(2\theta)}=\dfrac{1-\cos(2\theta)}{\sin(2\theta)}[/tex]
