If sin theta = -frac{7}{25} and theta is in t | vedclass

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(A) Given $\sin 	heta = -\frac{7}{25}$ and $	heta$ is in the third quadrant.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\cos^2 	heta = 1 - (

Vedclass · Accepted Answer

$\frac{17}{31}$

Vedclass · Answer

(A) Given $\sin 	heta = -\frac{7}{25}$ and $	heta$ is in the third quadrant.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\cos^2 	heta = 1 - (-\frac{7}{25})^2 = 1 - \frac{49}{625} = \frac{576}{625}$.Since $	heta$ is in the third quadrant,$\cos 	heta = -\frac{24}{25}$.Then $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{-7/25}{-24/25} = \frac{7}{24}$ and $\cot 	heta = \frac{1}{	an 	heta} = \frac{24}{7}$.Substituting these values into the expression:$\frac{7 \cot 	heta - 24 	an 	heta}{7 \cot 	heta + 24 	an 	heta} = \frac{7(\frac{24}{7}) - 24(\frac{7}{24})}{7(\frac{24}{7}) + 24(\frac{7}{24})}$$= \frac{24 - 7}{24 + 7} = \frac{17}{31}$.

If $\sin \theta = -\frac{7}{25}$ and $\theta$ is in the third quadrant,then $\frac{7 \cot \theta - 24 \tan \theta}{7 \cot \theta + 24 \tan \theta} = $

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