Question

The most general solution of $tan\theta =-1andcos\theta =\frac{1}{\sqrt{2}}$ 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. None of these

Answer 1

Answer: (C)

$2n\pi +\frac{7\pi }{4},nϵI$

Given:-

$\tan \theta =-1$

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

Solution:-

$\tan \theta =-1$

$\Rightarrow \tan \theta =-\tan \frac{\pi }{4}$

$\Rightarrow \tan \theta =\tan (2\pi -\frac{\pi }{4})$

$\Rightarrow \theta =n\pi +\frac{7\pi }{4}.....\left(1\right)$

Now,

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

$\Rightarrow \cos \theta =\cos \frac{\pi }{4}=\cos (-\frac{\pi }{4})$

$\Rightarrow \cos \theta =\cos (2\pi -\frac{\pi }{4})$

$\Rightarrow \theta =2n\pi +\frac{7\pi }{4}.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\theta =2n\pi +\frac{7\pi }{4};n\in I$

Hence the general solution of $\theta$ is $\theta =2n\pi +\frac{7\pi }{4}$.

Hence the correct answer is $\left(C\right)2n\pi +\frac{7\pi }{4};n\in I$.