Using the definitions of the corresponding functions, prove…

Using the definitions of the corresponding functions, prove: $\ \tan2z=\dfrac{2\tan z}{1-\tan^2z}$.

Step 1 · Double-angle forms of $\sin$ and $\cos$

Setting $z_1=z_2=z$ in the addition formulas (proved previously):

$$ \sin2z=\sin(z+z)=\sin z\cos z+\cos z\sin z=2\sin z\cos z $$

$$ \cos2z=\cos(z+z)=\cos z\cos z-\sin z\sin z=\cos^2z-\sin^2z $$

Step 2 · Form the ratio

$$ \tan2z=\frac{\sin2z}{\cos2z}=\frac{2\sin z\cos z}{\cos^2z-\sin^2z} $$

Step 3 · Divide through by $\cos^2z$

Valid wherever $\cos z\neq0$:

$$ \tan2z=\frac{\dfrac{2\sin z\cos z}{\cos^2z}}{\dfrac{\cos^2z-\sin^2z}{\cos^2z}}=\frac{2\dfrac{\sin z}{\cos z}}{1-\dfrac{\sin^2z}{\cos^2z}}=\frac{2\tan z}{1-\tan^2z} $$

Verification

Numerically at $z=0.7+1.3i$: direct evaluation gives $\tan(2z)\approx0.0037347+1.0104430i$, while $\dfrac{2\tan z}{1-\tan^2z}\approx0.0037347+1.0104430i$ — the two agree to full precision.