Question

Verify that tan 30$°$ = $\frac{\tan 60°-\tan 30°}{1+\tan 60°\tan 30°}$

Answer 1

Here, LHS = tan 30^o
= $\frac{1}{\sqrt{3}}(\mathrm{From}\mathrm{table})$


$$\begin{gathered} \mathrm{Now},\mathrm{RHS}=\frac{\tan 60°-\tan 30°}{1+\tan 60°\tan 30°} \\ =\frac{\sqrt{3}-{\displaystyle \frac{1}{\sqrt{3}}}}{1+\sqrt{3}\times {\displaystyle \frac{1}{\sqrt{3}}}}(\mathrm{From}\mathrm{table}) \\ =\frac{{\displaystyle \frac{3-1}{\sqrt{3}}}}{1+1} \\ =\frac{2}{2\sqrt{3}} \\ =\frac{1}{\sqrt{3}}=\mathrm{RHS} \end{gathered}$$