The most general value of theta which satisfi | vedclass

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

Question

(B) Given trigonometric equations are $	an 	heta = -1$ and $\cos 	heta = \frac{1}{\sqrt{2}}$.Since $	an 	heta$ is negative and $\cos 	heta$ is p

Vedclass · Accepted Answer

$2 n \pi + \frac{7 \pi}{4}$

Vedclass · Answer

(B) Given trigonometric equations are $	an 	heta = -1$ and $\cos 	heta = \frac{1}{\sqrt{2}}$.Since $	an 	heta$ is negative and $\cos 	heta$ is positive,$	heta$ must lie in the fourth quadrant.The general solution for $	an 	heta = -1$ is $	heta = n \pi + \frac{3 \pi}{4}$.The general solution for $\cos 	heta = \frac{1}{\sqrt{2}}$ is $	heta = 2 n \pi \pm \frac{\pi}{4}$.For $	heta$ to satisfy both,we look for the angle in the fourth quadrant where $\cos 	heta = \frac{1}{\sqrt{2}}$ and $	an 	heta = -1$,which is $	heta = 2 n \pi + \frac{7 \pi}{4}$ (or $2 n \pi - \frac{\pi}{4}$).

The most general value of $\theta$ which satisfies both the equations $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$ is

Similar Questions

If the angles of dip at two places are $30^o$ and $45^o$ respectively,then the ratio of horizontal components of earth's magnetic field at the two places will be

Which of the following substances releases nitrogen gas upon thermal decomposition?

If $I_n = \int \sin^n x \, dx$,then $n I_n - (n-1) I_{n-2}$ equals

If $\frac{1}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}$, then $C+D$ is equal to

Due to low ionization energy,alkali metals are .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module