Question

The most general solution of $\tan \theta =-1$ and $\cos \theta =\frac{1}{\sqrt{2}}$ is:

• A. $n\pi +\frac{7\pi }{4}$
• B. $n\pi +{(-1)}^n\frac{7\pi }{4}$
• C. $2n\pi +\frac{7\pi }{4}$
• D. none of these

Answer 1

Answer: (C)

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

Given, $\tan \theta =-1=\tan \frac{3\pi }{4}$

$\theta =n\pi +\frac{3\pi }{4}$

$\cos \theta =\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}$

$\theta =2n\pi \pm \frac{\pi }{4}$

Negative $\tan$ and positive $\cos$ is possible only in 4th quadrant so,principle solution $=\frac{7\pi }{4}$

$\therefore$ general solution is $2n\pi +\frac{7\pi }{4}$