Question

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

• A. $n\pi =\frac{7\pi }{4},nϵI$
• B. $n\pi +(-1{)}^n\frac{7\pi }{4},nϵI$
• C. $2n\pi =\frac{7\pi }{4},nϵI$
• D. $Noneofthese$

Answer 1

The correct option is A $n\pi =\frac{7\pi }{4},nϵI$

$tan\theta =-1$

$\Rightarrow$ $\theta =\frac{3\pi }{4}or\frac{7\pi }{4}$ (Since, tan has negative value in second and fourth quadrant)

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

$\Rightarrow$$\theta =\frac{\pi }{4}or\frac{7\pi }{4}$

$\therefore$ The general solution is $\frac{7\pi }{4}$.

So. correct option is 'A'.