question use the appropriate angle sum or difference formula to find the exact value of tan(-53π/12). answer…

Answer:

$2 + \sqrt{3}$

Explanation:

Step1: Rewrite the angle

$-\frac{53\pi}{12}= - \frac{48\pi + 5\pi}{12}=-4\pi-\frac{5\pi}{12}$ Since $\tan(x - 2k\pi)=\tan(x)$ for integer $k$, and here $k = 2$, so $\tan(-\frac{53\pi}{12})=\tan(-\frac{5\pi}{12})$. Also, $\tan(-\alpha)=-\tan(\alpha)$, then $\tan(-\frac{5\pi}{12})=-\tan(\frac{5\pi}{12})$.

Step2: Express $\frac{5\pi}{12}$ as a sum

$\frac{5\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6}$

Step3: Use the tangent - sum formula

The tangent - sum formula is $\tan(A + B)=\frac{\tan A+\tan B}{1 - \tan A\tan B}$. For $A=\frac{\pi}{4}$ and $B = \frac{\pi}{6}$, we know that $\tan\frac{\pi}{4}=1$ and $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$. $\tan(\frac{\pi}{4}+\frac{\pi}{6})=\frac{\tan\frac{\pi}{4}+\tan\frac{\pi}{6}}{1-\tan\frac{\pi}{4}\tan\frac{\pi}{6}}=\frac{1+\frac{\sqrt{3}}{3}}{1 - 1\times\frac{\sqrt{3}}{3}}$.

Step4: Simplify the expression

$\frac{1+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}=\frac{\frac{3 + \sqrt{3}}{3}}{\frac{3-\sqrt{3}}{3}}=\frac{3+\sqrt{3}}{3 - \sqrt{3}}$. Rationalize the denominator: $\frac{(3+\sqrt{3})(3 + \sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}=\frac{9 + 6\sqrt{3}+3}{9 - 3}=\frac{12 + 6\sqrt{3}}{6}=2+\sqrt{3}$. Since $\tan(-\frac{53\pi}{12})=-\tan(\frac{5\pi}{12})$, the value of $\tan(-\frac{53\pi}{12})=-(2 + \sqrt{3})=- 2-\sqrt{3}$.