If cos theta = - frac{1}{sqrt{2}} and tan the | vedclass

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

Question

(B) Given $\cos 	heta = -\frac{1}{\sqrt{2}}$ and $	an 	heta = 1$.Since $\cos 	heta$ is negative and $	an 	heta$ is positive,$	heta$ must lie in

Vedclass · Accepted Answer

$(2n + 1)\pi + \frac{\pi}{4}$

Vedclass · Answer

(B) Given $\cos 	heta = -\frac{1}{\sqrt{2}}$ and $	an 	heta = 1$.Since $\cos 	heta$ is negative and $	an 	heta$ is positive,$	heta$ must lie in the third quadrant.The principal value of $	heta$ in the third quadrant is $	heta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$.The general solution for $	heta$ when $\cos 	heta = \cos \alpha$ and $\sin 	heta = \sin \alpha$ is $	heta = 2n\pi + \alpha$.Thus,the general value is $	heta = 2n\pi + \frac{5\pi}{4}$.Since $\frac{5\pi}{4} = \pi + \frac{\pi}{4}$,we can write this as $	heta = 2n\pi + \pi + \frac{\pi}{4} = (2n + 1)\pi + \frac{\pi}{4}$.

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

Similar Questions

If $2\sin^2 \theta = 3\cos \theta$,where $0 \le \theta \le 2\pi$,then $\theta = $

The number of solutions of the equation $\cos 2 \theta \cos \frac{\theta}{2} + \cos \frac{5 \theta}{2} = 2 \cos^3 \frac{5 \theta}{2}$ in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is:

What is the number of roots of the quadratic equation $8\sec^2\theta - 6\sec\theta + 1 = 0$?

The number of values of $x$ in the interval $\left(\frac{\pi}{4}, \frac{7 \pi}{4}\right)$ for which $14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 21 - 4 \cos^{2} x$ holds,is

The set of solutions of the equation $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module