Question

If $\cos \theta =-\frac{1}{\sqrt{2}}$ and $\tan \theta =1$, then the general value of $\theta$ is

• A. $2n\pi +\frac{\pi }{4},n=0,1,2,3.\dots \dots$.
• B. $(2n+1)\pi +\frac{\pi }{4},n=0,1,2,3\dots \dots \dots$
• C. $n\pi +\frac{\pi }{4},n=0,1,2,3\dots \dots \dots$
• D. $n\pi \pm \frac{\pi }{4},n=0,1,2,3\dots \dots$

Answer 1

Answer: (B)

$\cos \theta =\frac{-1}{\sqrt{2}}$ and $\tan \theta =1$
$\Rightarrow \cos \theta =-\cos \frac{\pi }{4}$ and $\tan \theta =\tan \frac{\pi }{4}$
$\Rightarrow \cos \theta =\cos \left(\pi +\frac{\pi }{4}\right)$ and $\tan \theta =\tan \left(\pi +\frac{\pi }{4}\right)$
[ $\because$ in IIIrd quadrant, cosine is negative and tangent is positive]
$\Rightarrow \theta =(2n+1)\pi +\frac{\pi }{4},n=0,1,2,3,\dots$