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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 = $

Similar Questions

Find the radian measure corresponding to the following degree measure: $520^{\circ}$

$\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \dots \cos 179^{\circ} = $

$\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ \dots \tan 89^\circ = $

If $\tan \theta = 2$ and $\theta$ lies in the third quadrant,then the value of $\sec \theta$ is

The expression $\frac{\tan \left( \frac{3\pi}{2} - \alpha \right) \cos \left( \frac{3\pi}{2} - \alpha \right)}{\cos (2\pi - \alpha )} + \cos \left( \alpha - \frac{\pi}{2} \right) \sin (\pi - \alpha ) + \cos (\pi + \alpha ) \sin \left( \alpha - \frac{\pi}{2} \right)$ when simplified reduces to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module