If tan theta = - frac{1}{sqrt{10}} and theta | vedclass

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(C) Given $	an 	heta = -\frac{1}{\sqrt{10}}$.Since $	heta$ lies in the fourth quadrant,$\cos 	heta$ must be positive.We use the identity $1 + 	an

Vedclass · Accepted Answer

$\sqrt{\frac{10}{11}}$

Vedclass · Answer

(C) Given $	an 	heta = -\frac{1}{\sqrt{10}}$.Since $	heta$ lies in the fourth quadrant,$\cos 	heta$ must be positive.We use the identity $1 + 	an^2 	heta = \sec^2 	heta$.Substituting the value of $	an 	heta$:$1 + \left(-\frac{1}{\sqrt{10}}ight)^2 = \sec^2 	heta$$1 + \frac{1}{10} = \sec^2 	heta$$\sec^2 	heta = \frac{11}{10}$Since $\cos 	heta = \frac{1}{\sec 	heta}$,we have $\cos^2 	heta = \frac{1}{\sec^2 	heta} = \frac{10}{11}$.Since $	heta$ is in the fourth quadrant,$\cos 	heta$ is positive,so $\cos 	heta = \sqrt{\frac{10}{11}}$.

If $\tan \theta = - \frac{1}{\sqrt{10}}$ and $\theta$ lies in the fourth quadrant,then $\cos \theta = $

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