Prove the identity: $\left(1 + \tan \beta \tan 2 \beta\right) \sin 2 \beta = \tan 2 \beta$
To prove that (TPT): $\;\;\;$ $\left(1 + \tan \beta \tan 2 \beta\right) \sin 2 \beta = \tan 2 \beta$
i.e. $\;$ TPT $\;\;\;$ $1 + \tan \beta \tan 2 \beta = \dfrac{\tan 2 \beta}{\sin 2 \beta}$
i.e. $\;$ TPT $\;\;\;$ $1 + \tan \beta \tan 2 \beta = \dfrac{\sin 2 \beta}{\cos 2 \beta} \times \dfrac{1}{\sin 2\beta}$
i.e. $\;$ TPT $\;\;\;$ $\left(1 + \tan \beta \tan 2 \beta\right) \cos 2 \beta = 1$
$\begin{aligned}
LHS & = \left(1 + \tan \beta \tan 2 \beta\right) \cos 2 \beta \\\\
& = \cos 2 \beta + \dfrac{\sin \beta}{\cos \beta} \times \dfrac{\sin 2 \beta}{\cos 2 \beta} \times \cos 2 \beta \\\\
& = \cos 2 \beta + \dfrac{\sin \beta 2 \sin \beta \cos \beta}{\cos \beta} \\\\
& = \cos^2 \beta - \sin^2 \beta + 2 \sin^2 \beta \\\\
& = \cos^2 \beta + \sin^2 \beta \\\\
& = 1 = RHS
\end{aligned}$
Hence proved.