It is required to prove that:
[tex]\frac{\tan x+1}{\tan x-1}=\frac{1+2\sin x\cos x}{\sin^2x-\cos^2x}[/tex]Solve the right hand side as follows:
[tex]\begin{gathered} \frac{1+2\sin x\cos x}{\sin^2x-\cos^2x}=\frac{\sin^2x+\cos^2x+2\sin x\cos x}{(\sin x-\cos x)\cdot(\sin x+\cos x)_{}} \\ =\frac{(\sin x+\cos x)^2}{(\sin x-\cos x)\cdot(\sin x+\cos x)} \\ =\frac{\sin x+\cos x}{\sin x-\cos x} \end{gathered}[/tex]Divide the numerator and denominator by cosx to get:
[tex]\begin{gathered} \frac{\sin x+\cos x}{\sin x-\cos x}=\frac{\frac{\sin x}{\cos x}+\frac{\cos x}{cosx}}{\frac{\sin x}{\cos x}-\frac{\cos x}{cosx}} \\ =\frac{\tan x+1}{\text{tanx}-1} \end{gathered}[/tex]Which is the left side of the equation.
Hence it is proved that:
[tex]\frac{\tan x+1}{\tan x-1}=\frac{1+2\sin x\cos x}{\sin^2x-\cos^2x}[/tex]