Question

write the following expression as the sine, cosine, or tangent of a double - angle. then find the expression.

\\( \frac { 2 \tan \frac { \pi } { 6 } } { 1 - \tan ^ { 2 } \frac { \pi } { 6 } } \\)

b. \\( \frac { 2 \tan \frac { \pi } { 6 } } { 1 - \tan ^ { 2 } \frac { \pi } { 6 } } = \tan \frac { \pi } { 3 } \\)

c. \\( \frac { 2 \tan \frac { \pi } { 6 } } { 1 - \tan ^ { 2 } \frac { \pi } { 6 } } = \cos \\)

the exact value of the expression is
(type an exact answer, using radicals as needed. rationalize the denominator.)

Explanation:

Step1: Recall the double - angle formula for tangent

The double - angle formula for tangent is \(\tan(2\alpha)=\frac{2\tan\alpha}{1 - \tan^{2}\alpha}\).
In the given expression \(\frac{2\tan\frac{\pi}{6}}{1-\tan^{2}\frac{\pi}{6}}\), if we let \(\alpha=\frac{\pi}{6}\), then by the formula \(\frac{2\tan\frac{\pi}{6}}{1-\tan^{2}\frac{\pi}{6}}=\tan(2\times\frac{\pi}{6})=\tan\frac{\pi}{3}\).

Step2: Calculate the value of \(\tan\frac{\pi}{3}\)

We know that \(\tan\frac{\pi}{3}=\sqrt{3}\).

Answer:

\(\sqrt{3}\)