If sin theta = frac{3}{5} and theta is not in | vedclass

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(D) Given: $\sin 	heta = \frac{3}{5}$.Since $	heta$ is not in the first quadrant and $\sin 	heta$ is positive,$	heta$ must be in the second quadra

Vedclass · Accepted Answer

$4$

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(D) Given: $\sin 	heta = \frac{3}{5}$.Since $	heta$ is not in the first quadrant and $\sin 	heta$ is positive,$	heta$ must be in the second quadrant.Therefore,$\cos 	heta = -\sqrt{1 - \sin^2 	heta} = -\sqrt{1 - \frac{9}{25}} = -\frac{4}{5}$.Now,$\sin 2 	heta = 2 \sin 	heta \cos 	heta = 2 \left(\frac{3}{5}ight) \left(-\frac{4}{5}ight) = -\frac{24}{25}$.$\cos 2 	heta = 1 - 2 \sin^2 	heta = 1 - 2 \left(\frac{9}{25}ight) = 1 - \frac{18}{25} = \frac{7}{25}$.$	an 2 	heta = \frac{\sin 2 	heta}{\cos 2 	heta} = \frac{-24/25}{7/25} = -\frac{24}{7}$.Substituting these values into the expression:$15 \sin 2 	heta - 20 \cos 2 	heta - 7 	an 2 	heta = 15 \left(-\frac{24}{25}ight) - 20 \left(\frac{7}{25}ight) - 7 \left(-\frac{24}{7}ight)$.$= -\frac{72}{5} - \frac{28}{5} + 24 = -\frac{100}{5} + 24 = -20 + 24 = 4$.

If $\sin \theta = \frac{3}{5}$ and $\theta$ is not in the first quadrant,then $15 \sin 2 \theta - 20 \cos 2 \theta - 7 \tan 2 \theta = $

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