If tan theta = - frac{1}{sqrt{3}},sin theta = | vedclass

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

Question

(B) Given that $	an 	heta = -\frac{1}{\sqrt{3}}$,$\sin 	heta = \frac{1}{2}$,and $\cos 	heta = -\frac{\sqrt{3}}{2}$.Since $\sin 	heta > 0$ and $\c

Vedclass · Accepted Answer

$\frac{5\pi}{6}$

Vedclass · Answer

(B) Given that $	an 	heta = -\frac{1}{\sqrt{3}}$,$\sin 	heta = \frac{1}{2}$,and $\cos 	heta = -\frac{\sqrt{3}}{2}$.Since $\sin 	heta > 0$ and $\cos 	heta The reference angle $\alpha$ is given by $	an \alpha = \frac{1}{\sqrt{3}}$,which implies $\alpha = \frac{\pi}{6}$.In the second quadrant,$	heta = \pi - \alpha = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.Thus,the principal value of $	heta$ is $\frac{5\pi}{6}$.

If $\tan \theta = - \frac{1}{\sqrt{3}}$,$\sin \theta = \frac{1}{2}$,and $\cos \theta = - \frac{\sqrt{3}}{2}$,then the principal value of $\theta$ is:

Similar Questions

The interval for which ${\sin ^{ - 1}}\sqrt x + {\cos ^{ - 1}}\sqrt x = \frac{\pi }{2}$ holds is:

The trigonometric equation $\sin ^{-1} x = 2 \sin ^{-1} a$ has a solution

$\tan \left[ \frac{1}{2} \cos^{-1} \left( \frac{\sqrt{5}}{3} \right) \right] = $

If $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$,where $0 < x < 1$ and $0 < y < \frac{\pi}{2}$,then $\frac{dy}{dx} = $

If $y = \tan^{-1} \left[ \sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} \right]$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module