Question

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

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

Answer 1

The correct option is C $2n\pi +\frac{7\pi }{4},n\in I$

$\tan \theta =-1\Rightarrow \theta =n\pi -\frac{\pi }{4}$ ...(1)
$\cos \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =2n\pi \pm \frac{\pi }{4}$ ...(2)
From (1) and (2), we get
$\theta =2n\pi +\frac{7\pi }{4}$