The most general value of theta which satisfi | vedclass

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(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

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