Question

simplify the expression by using a double - angle formula or a half - angle formula.
(a) $\frac{2\tan(6^{circ})}{1 - \tan^{2}(6^{circ})}$
(b) $\frac{2\tan(6\theta)}{1 - \tan^{2}(6\theta)}$

Explanation:

Step1: Recall double - angle formula for tangent

The double - angle formula for tangent is $\tan(2\alpha)=\frac{2\tan\alpha}{1 - \tan^{2}\alpha}$.

Step2: Solve part (a)

For the expression $\frac{2\tan(6^{\circ})}{1 - \tan^{2}(6^{\circ})}$, comparing with the double - angle formula $\tan(2\alpha)=\frac{2\tan\alpha}{1 - \tan^{2}\alpha}$, where $\alpha = 6^{\circ}$, we get $\tan(2\times6^{\circ})=\tan(12^{\circ})$.

Step3: Solve part (b)

For the expression $\frac{2\tan(6\theta)}{1 - \tan^{2}(6\theta)}$, comparing with the double - angle formula $\tan(2\alpha)=\frac{2\tan\alpha}{1 - \tan^{2}\alpha}$, where $\alpha = 6\theta$, we get $\tan(2\times6\theta)=\tan(12\theta)$.

Answer:

(a) $\tan(12^{\circ})$
(b) $\tan(12\theta)$