Question

use a sum or difference identity to find the exact value. tan(-15°) a. (√3 - 4)/2 b. (√3 + 4)/2 c. √3 - 2 d. -√3 + 2

Explanation:

Step1: Use the tangent difference identity

The tangent difference identity is $\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$. Also, $\tan(-\alpha)=-\tan\alpha$, so $\tan(-15^{\circ})=-\tan15^{\circ}$. And $15^{\circ}=45^{\circ}-30^{\circ}$. Then $\tan15^{\circ}=\tan(45^{\circ}-30^{\circ})=\frac{\tan45^{\circ}-\tan30^{\circ}}{1+\tan45^{\circ}\tan30^{\circ}}$.
Since $\tan45^{\circ} = 1$ and $\tan30^{\circ}=\frac{\sqrt{3}}{3}$, we have $\tan(45^{\circ}-30^{\circ})=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}$.

Step2: Simplify the fraction

First, simplify the numerator and denominator:
The numerator is $\frac{3-\sqrt{3}}{3}$, and the denominator is $\frac{3 + \sqrt{3}}{3}$. So $\tan(45^{\circ}-30^{\circ})=\frac{3-\sqrt{3}}{3+\sqrt{3}}$.
Rationalize the denominator: multiply the numerator and denominator by $3-\sqrt{3}$.

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Since $\tan(-15^{\circ})=-\tan15^{\circ}$, then $\tan(-15^{\circ})=-(2 - \sqrt{3})=\sqrt{3}-2$.

Answer:

C. $\sqrt{3}-2$