If tan theta = -frac{4}{3}, then sin theta = | vedclass

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(B) Given $	an 	heta = -\frac{4}{3}.$ We know that $\sec^2 	heta = 1 + 	an^2 	heta = 1 + \left(-\frac{4}{3}ight)^2 = 1 + \frac{16}{9} = \frac{2

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$-4/5$ or $4/5$

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(B) Given $	an 	heta = -\frac{4}{3}.$ We know that $\sec^2 	heta = 1 + 	an^2 	heta = 1 + \left(-\frac{4}{3}ight)^2 = 1 + \frac{16}{9} = \frac{25}{9}.$ Thus,$\cos^2 	heta = \frac{1}{\sec^2 	heta} = \frac{9}{25}.$ Using $\sin^2 	heta = 1 - \cos^2 	heta,$ we get $\sin^2 	heta = 1 - \frac{9}{25} = \frac{16}{25}.$ Therefore,$\sin 	heta = \pm \frac{4}{5}.$ Since $	an 	heta$ is negative,$	heta$ lies in the $2^{nd}$ or $4^{th}$ quadrant. In the $2^{nd}$ quadrant,$\sin 	heta$ is positive $(4/5)$,and in the $4^{th}$ quadrant,$\sin 	heta$ is negative $(-4/5)$. Thus,both values are possible.

If $\tan \theta = -\frac{4}{3},$ then $\sin \theta = $

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