If theta lies in the third quadrant and cos t | vedclass

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(D) Given that $	heta$ lies in the third quadrant,where $	an 	heta$ is positive.We know that $\sin^2 	heta + \cos^2 	heta = 1$.Substituting $\cos

Vedclass · Accepted Answer

$\frac{4}{3}$

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(D) Given that $	heta$ lies in the third quadrant,where $	an 	heta$ is positive.We know that $\sin^2 	heta + \cos^2 	heta = 1$.Substituting $\cos 	heta = -\frac{3}{5}$,we get $\sin^2 	heta + (-\frac{3}{5})^2 = 1$.$\sin^2 	heta + \frac{9}{25} = 1 \implies \sin^2 	heta = 1 - \frac{9}{25} = \frac{16}{25}$.Since $	heta$ is in the third quadrant,$\sin 	heta$ must be negative,so $\sin 	heta = -\frac{4}{5}$.Now,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{-4/5}{-3/5} = \frac{4}{3}$.Hence,option $D$ is correct.

If $\theta$ lies in the third quadrant and $\cos \theta = -\frac{3}{5}$,find the value of $\tan \theta$.

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