If sin theta = frac{24}{25} and theta lies in | vedclass

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(C) Given $\sin 	heta = \frac{24}{25}$.Since $	heta$ lies in the second quadrant,$\cos 	heta$ and $	an 	heta$ are negative.Using the identity $\c

Vedclass · Accepted Answer

$-7$

Vedclass · Answer

(C) Given $\sin 	heta = \frac{24}{25}$.Since $	heta$ lies in the second quadrant,$\cos 	heta$ and $	an 	heta$ are negative.Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get $\cos^2 	heta = 1 - (\frac{24}{25})^2 = 1 - \frac{576}{625} = \frac{49}{625}$.Since $	heta$ is in the second quadrant,$\cos 	heta = -\sqrt{\frac{49}{625}} = -\frac{7}{25}$.Then,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{24/25}{-7/25} = -\frac{24}{7}$.Also,$\sec 	heta = \frac{1}{\cos 	heta} = -\frac{25}{7}$.Therefore,$\sec 	heta + 	an 	heta = -\frac{25}{7} + (-\frac{24}{7}) = -\frac{49}{7} = -7$.

If $\sin \theta = \frac{24}{25}$ and $\theta$ lies in the second quadrant,then $\sec \theta + \tan \theta = $

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