Find the value of tan frac{13 pi}{12}. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We have $	an \frac{13\pi}{12} = 	an \left( \pi + \frac{\pi}{12} ight)$.Since $	an(\pi + 	heta) = 	an 	heta$,we get $	an \frac{\pi}{12}$.N

Vedclass · Accepted Answer

$2 - \sqrt{3}$

Vedclass · Answer

(A) We have $	an \frac{13\pi}{12} = 	an \left( \pi + \frac{\pi}{12} ight)$.Since $	an(\pi + 	heta) = 	an 	heta$,we get $	an \frac{\pi}{12}$.Now,$\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6} = 15^\circ$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$:$	an \left( \frac{\pi}{4} - \frac{\pi}{6} ight) = \frac{	an \frac{\pi}{4} - 	an \frac{\pi}{6}}{1 + 	an \frac{\pi}{4} 	an \frac{\pi}{6}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$.Rationalizing the denominator:$\frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.

Find the value of $\tan \frac{13 \pi}{12}$.

Similar Questions

If two acute angles $A$ and $B$ are such that $A \neq B$ and $\frac{x}{y}=\frac{\cos A}{\cos B}$,then $\frac{x \tan A-y \tan B}{x+y}=$

If $\cos (A-B)=3/5$ and $\tan A \tan B=2$,then which one of the following is true?

The value of $\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}$ is

The value of $\cos \left(\frac{7 \pi}{12}\right)$ is

If $(1 + \tan \theta )(1 + \tan \phi ) = 2$,then $\theta + \phi = \dots ^\circ$

Vedclass Test Series

Exam Paper Generator

Online Exam Module