Solution assignment 09 Addition and subtraction formulas

Show:

$\tan(2x)=\displaystyle\frac{2\tan(x)}{1-\tan^2(x)}$

In the right-hand side we substitute:

$\tan(x)=\displaystyle\frac{\sin(x)}{\cos(x)}$

and get:

$\displaystyle\frac{2\displaystyle\frac{\sin(x)}{\cos(x)}}{1-\displaystyle\frac{\sin^2(x)}{\cos^2(x)}}$

We multiply both numerator and denominator by $\cos^2(x)$ :

$\displaystyle\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\displaystyle\frac{\sin(2x)}{\cos(2x)}=\tan(2x)$