The value of tan frac{2pi}{5} - tan frac{pi}{ | vedclass

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Question

(D) We know the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Let $A = \frac{2\pi}{5} = \frac{6\pi}{15}$ and $B = \frac{\pi}{15}$.

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(D) We know the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Let $A = \frac{2\pi}{5} = \frac{6\pi}{15}$ and $B = \frac{\pi}{15}$.Then $A - B = \frac{6\pi}{15} - \frac{\pi}{15} = \frac{5\pi}{15} = \frac{\pi}{3}$.Using the formula,$	an\left(\frac{6\pi}{15} - \frac{\pi}{15}ight) = \frac{	an \frac{6\pi}{15} - 	an \frac{\pi}{15}}{1 + 	an \frac{6\pi}{15} 	an \frac{\pi}{15}} = 	an \frac{\pi}{3}$.Since $	an \frac{\pi}{3} = \sqrt{3}$,we have $\frac{	an \frac{2\pi}{5} - 	an \frac{\pi}{15}}{1 + 	an \frac{2\pi}{5} 	an \frac{\pi}{15}} = \sqrt{3}$.Multiplying both sides by the denominator: $	an \frac{2\pi}{5} - 	an \frac{\pi}{15} = \sqrt{3} (1 + 	an \frac{2\pi}{5} 	an \frac{\pi}{15})$.$	an \frac{2\pi}{5} - 	an \frac{\pi}{15} = \sqrt{3} + \sqrt{3} 	an \frac{2\pi}{5} 	an \frac{\pi}{15}$.Rearranging the terms: $	an \frac{2\pi}{5} - 	an \frac{\pi}{15} - \sqrt{3} 	an \frac{2\pi}{5} 	an \frac{\pi}{15} = \sqrt{3}$.

The value of $\tan \frac{2\pi}{5} - \tan \frac{\pi}{15} - \sqrt{3} \tan \frac{2\pi}{5} \tan \frac{\pi}{15}$ is equal to

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