Question

calculate the value f tan15

Answer 1

$$\begin{gathered} \tan 15°=\tan \left(45°-30°\right) \\ \mathrm{now},\tan \left(\mathrm{A}-\mathrm{B}\right)=\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A}\tan \mathrm{B}} \\ \mathrm{so}, \\ \tan 15°=\frac{\tan 45°-\tan 30°}{1+\tan 45°\tan 30°} \\ \Rightarrow \tan 15°=\frac{1-{\displaystyle \frac{1}{\sqrt{3}}}}{1+{\displaystyle \frac{1}{\sqrt{3}}}} \\ \Rightarrow \tan 15°=\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ \Rightarrow \tan 15°=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1} \\ \Rightarrow \tan 15°=\frac{{\left(\sqrt{3}-1\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2}=\frac{3+1-2\sqrt{3}}{3-1} \\ \Rightarrow \tan 15°=\frac{4-2\sqrt{3}}{2}=\frac{2\left(2-\sqrt{3}\right)}{2} \\ \Rightarrow \tan 15°=2-\sqrt{3} \end{gathered}$$