Question

The most general solution of $tan\theta =-1,cos\theta =1/\sqrt{2}$ is-

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

Answer 1

The correct option is C $2n\pi +7\pi /4$

Given $tan\theta =-1$
$tan\theta =-\tan \frac{\pi }{4}$
$tan\theta =\tan (\pi -\frac{\pi }{4})ortan\theta =\tan (2\pi -\frac{\pi }{4})$
$\Rightarrow \tan \theta =\tan \frac{3\pi }{4}$ or $\tan \theta =\tan \frac{7\pi }{4}$
$\Rightarrow \theta =n\pi +\alpha$ where $\alpha =\frac{3\pi }{4}or\frac{7\pi }{4}$
Also, given $\cos \theta =1/\sqrt{2}$
$\cos \theta =\cos \frac{\pi }{4}or\cos \theta =\cos \left(2\pi -\frac{\pi }{4}\right)$
$\Rightarrow \cos \theta =\cos \frac{\pi }{4}or\cos \theta =\cos \frac{7\pi }{4}$
$\Rightarrow \theta =2n\pi \pm \alpha$ where $\alpha =\frac{\pi }{4}or\frac{7\pi }{4}$
Hence, $\alpha =\frac{7\pi }{4}$ is the common solution satisfying both.