Question

The most general value of $\theta$ satisfying the equations $\tan \theta =-1$ 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

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

$\begin{array}{c}\tan \theta =-1=\tan \frac{3\pi }{4}=\tan \frac{7\pi }{4}\\ \tan \left(\pi -\frac{3\pi }{4}\right)=\tan \left(\pi +\frac{3\pi }{4}\right)\\ \cos \theta =\frac{1}{\sqrt{2}}\\ \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\\ \cos \left(2\pi -\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}=\cos \frac{7\pi }{4}\\ \therefore generalvalueis2n\pi +\frac{7\pi }{4}\end{array}$

Hence, Option $C$ is correct .