If x = 30°, verify that tan 2x = $\dfrac{\text{2 tan x}}{\text{1 - tan}^2 x}$.

To verify,

tan 2x = $\dfrac{\text{2 tan x}}{\text{1 - tan}^2 x}$.

Substituting value of x in L.H.S. of the above equation,

⇒ tan 2x = tan 2(30°) = tan 60° = $\sqrt{3}$.

Substituting value of x in R.H.S. of the equation.

$\Rightarrow \dfrac{\text{2 tan x}}{\text{1 - tan}^2 x} \\[1em] \Rightarrow \dfrac{2 \times \text{tan 30°}}{1 - \text{tan}^2 30°} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 - \Big(\dfrac{1}{\sqrt{3}}\Big)^2} \\[1em] \Rightarrow \dfrac{\dfrac{2}{\sqrt{3}}}{1 - \dfrac{1}{3}} \\[1em] \Rightarrow \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} \\[1em] \Rightarrow \dfrac{2 \times 3}{2 \times \sqrt{3}} \\[1em] \Rightarrow \sqrt{3}.$

Since, L.H.S. = R.H.S.

Hence, proved that tan 2x = $\dfrac{\text{2 tan x}}{\text{1 - tan}^2 x}$.