Question

Find the most general value of $\theta$ satisfying the equation $\tan \theta =-1$ and $\cos \theta =\frac{1}{\sqrt{2}}$.

Answer 1

$tan\theta =-1$ lies in IV quadrant

$cos\theta =\frac{1}{\sqrt{2}}$ lies in I & IV quadrant

$\therefore \theta =\frac{7\pi }{4}+2n\pi ,n\in Z$