Question

use a sum or difference identity to find the exact value.

\\( \frac { \tan \frac { \pi } { 15 } + \tan \frac { 4 \pi } { 15 } } { 1 - \tan \frac { \pi } { 15 } \tan \frac { 4 \pi } { 15 } } \\)

a. \\( \sqrt { 3 } \\)
b. \\( \frac { \sqrt { 3 } } { 2 } \\)
c. \\( \frac { 1 } { 2 } \\)
d. \\( \frac { \sqrt { 3 } } { 3 } \\)

Explanation:

Step1: Recall the tangent sum identity

The tangent sum identity is \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\).
Let \(A=\frac{\pi}{15}\) and \(B = \frac{4\pi}{15}\).

Step2: Calculate \(A + B\)

\(A + B=\frac{\pi}{15}+\frac{4\pi}{15}=\frac{\pi + 4\pi}{15}=\frac{5\pi}{15}=\frac{\pi}{3}\).

Step3: Find the value of \(\tan(A + B)\)

Since \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\), and \(A + B=\frac{\pi}{3}\), then \(\frac{\tan\frac{\pi}{15}+\tan\frac{4\pi}{15}}{1-\tan\frac{\pi}{15}\tan\frac{4\pi}{15}}=\tan(\frac{\pi}{3})\).
We know that \(\tan(\frac{\pi}{3})=\sqrt{3}\).

Answer:

A. \(\sqrt{3}\)