The problem stated that

whereas sin 15 and cos 15 can be evaluated using half-angle formula. The half-angle formula for sine function is
![\sin (\theta)/(2)=\sqrt[]{(1-\cos \theta)/(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/k4eai17biw10bwwpv4nkisi7s50s5fdsfb.png)
while the half-angle formula for cosine function is
![\cos (\theta)/(2)=\sqrt[]{(1+\cos \theta)/(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/khnfj9qzjeg9qr9q2lkhoy59ahcgvd63tj.png)
The theta value to evaluate the half-angle formula for tangent 15 is 30 degrees. We now proceed in calculating the value of the trigonometric function.
![\begin{gathered} \sin (30)/(2)=\sqrt[]{(1-\cos30)/(2)} \\ \cos (30)/(2)=\sqrt[]{(1+\cos30)/(2)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/769yp8zvl958zh1kruua3lnws9wsbadflz.png)
The exact value of cosine 30 is √3/2, hence, the equation above now becomes
![\begin{gathered} \sin (30)/(2)=\sqrt[]{\frac{1-\frac{\sqrt[]{3}}{2}}{2}} \\ \cos (30)/(2)=\sqrt[]{\frac{1+\frac{\sqrt[]{3}}{2}}{2}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qhl28o4g1qlowt9jfudvz7bt64ub4d8uvn.png)
Simplifying the equations above
![\begin{gathered} \sin (30)/(2)=\sqrt[]{\frac{\frac{2-\sqrt[]{3}}{2}}{2}}=\sqrt[]{\frac{2-\sqrt[]{3}}{4}} \\ \cos (30)/(2)=\sqrt[]{\frac{\frac{2+\sqrt[]{3}}{2}}{2}=}\sqrt[]{\frac{2+\sqrt[]{3}}{4}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8s5hry0du3mx5960kj330u4vvqirgprdcl.png)
Since tangent theta is the ratio between the sin function and the cosine function, getting the ratio of the two equations above will result to
![(\sin(30)/(2))/(\cos(30)/(2))=\tan 15=\frac{\sqrt[]{\frac{2-\sqrt[]{3}}{4}}}{\sqrt[]{\frac{2+\sqrt[]{3}}{4}}}](https://img.qammunity.org/2023/formulas/mathematics/college/vqzfkorornmn66ywt4hh84sjq8sk3fahpi.png)
Simplifying,
![\begin{gathered} \tan 15=\sqrt[]{\frac{2-\sqrt[]{3}}{4}}\cdot\sqrt[]{\frac{4}{2+\sqrt[]{3}}} \\ \tan 15=\frac{\sqrt[]{2-\sqrt[]{3}}}{\sqrt[]{2+\sqrt[]{3}}}=\sqrt[]{\frac{2-\sqrt[]{3}}{2+\sqrt[]{3}}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/m5lnnwjyxnzzk6cfeksabzkkmkpla8zl3u.png)
Where the equation above shows the exact value of tan 15.