Identity:
[tex]\frac{1-\cos2x}{\tan^2x}=1+\cos 2x[/tex]We can also write the given identity as :
[tex]\frac{1-\cos 2x}{1+\cos 2x}=\tan ^2x[/tex]From the trignometric ratio of tanx we have:
[tex]\tan x=\frac{\sin x}{\cos x}[/tex]and
[tex]\begin{gathered} \text{Cos}2x=1-2\sin ^2x \\ \Rightarrow1-\cos 2x=2\sin ^2x \\ \text{Cos}2x=2\cos ^2x-1 \\ \Rightarrow1+\cos 2x=2\cos ^2x \end{gathered}[/tex]So, substitute the value and simplify:
[tex]\begin{gathered} \frac{1-\cos2x}{1+\cos2x}=\tan ^2x \\ \frac{2\sin^2x}{2\cos^2x}=\tan ^2x \\ \frac{\sin ^2x}{\cos ^2x}=\tan ^2x \\ \text{tan}^2x=\tan ^2x \end{gathered}[/tex]So, LHS = RHS, the identity is proved
