ANSWER
2 - √3
EXPLANATION
15 degrees is the difference between 45 degrees and 30 degrees. We know the exact value of the tangent of these angles,
[tex]\begin{cases}\tan 30\degree=\frac{\sqrt[]{3}}{3} \\ \\ \tan 45\degree=1\end{cases}[/tex]
Using the identity,
[tex]\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}[/tex]
If A is 45° and B is 30°,
[tex]\tan (45-30)=\tan (15)=\frac{\tan45-\tan30}{1+\tan45\tan30}=\frac{1-\frac{\sqrt[]{3}}{3}}{1+1\cdot\frac{\sqrt[]{3}}{3}}=\frac{1-\frac{\sqrt[]{3}}{3}}{1+\frac{\sqrt[]{3}}{3}}[/tex]
Add the numbers in the denominator and the numerator,
[tex]\tan (15)=\frac{\frac{3-\sqrt[]{3}}{3}}{\frac{3+\sqrt[]{3}}{3}}=\frac{3-\sqrt[]{3}}{3+\sqrt[]{3}}[/tex]
To simplify this expression, multiply and divide by (3 - √3),
[tex]\tan (15)=\frac{(3-\sqrt[]{3})}{(3+\sqrt[]{3})}\cdot\frac{(3-\sqrt[]{3})}{(3-\sqrt[]{3})}[/tex]
The denominator is a difference of two squares, and the numerator is the square of the binomial,
[tex]\tan (15)=\frac{(3-\sqrt[]{3})^2}{3^2-(\sqrt[]{3})^2}=\frac{3^2-6\sqrt[]{3}+3}{9-3}=\frac{12-6\sqrt[]{3}}{6}=\frac{12}{6}-\frac{6\sqrt[]{3}}{6}=2-\sqrt[]{3}[/tex]
Hence, the value of tan15° is 2 - √3.
