If tan A

Question

(B) Given $	an 2A = \frac{2 	an A}{1 - 	an^2 A} = -\frac{4}{3}$.$6 	an A = -4 + 4 	an^2 A \Rightarrow 2 	an^2 A - 3 	an A - 2 = 0$.Solving for

Vedclass · Accepted Answer

$-\frac{117}{125}$

Vedclass · Answer

(B) Given $	an 2A = \frac{2 	an A}{1 - 	an^2 A} = -\frac{4}{3}$.$6 	an A = -4 + 4 	an^2 A \Rightarrow 2 	an^2 A - 3 	an A - 2 = 0$.Solving for $	an A$: $(2 	an A + 1)(	an A - 2) = 0$,so $	an A = -\frac{1}{2}$ or $	an A = 2$.Since $	an A Now,$\cos 2A = \frac{1 - 	an^2 A}{1 + 	an^2 A} = \frac{1 - 1/4}{1 + 1/4} = \frac{3/4}{5/4} = \frac{3}{5}$.Using the identity $\cos 6A = 4 \cos^3 2A - 3 \cos 2A$:$\cos 6A = 4 \left(\frac{3}{5}ight)^3 - 3 \left(\frac{3}{5}ight) = 4 \left(\frac{27}{125}ight) - \frac{9}{5} = \frac{108}{125} - \frac{225}{125} = -\frac{117}{125}$.

If $\tan A < 0$ and $\tan 2A = -\frac{4}{3}$,then $\cos 6A =$

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