Trigonometric functions and formulas

Question

Given that $\csc(\theta) = −\sqrt{2}$, $\tan{\theta} = 1$ and $−\pi < θ < \pi$, find the exact value of the angle $\theta$ in radians.

Answer 1

$\displaystyle\tan\theta=1\implies \theta=n\pi+\frac\pi4$ where $n$ is any integer

Now, $\displaystyle\csc\theta=-\sqrt2\iff \sin\theta=-\frac1{\sqrt2}$

Again, $\displaystyle\sin\left(n\pi+\frac\pi4\right)=\cos n\pi\sin\frac\pi4=(-1)^n\frac1{\sqrt2}$ as $\cos2r\pi=1,\cos(2s+1)\pi=-1$ for integer $r,s$

$\displaystyle\implies(-1)^n=-1\iff n$ must be odd $=2m+1$(say) where $m$ is any integer

$\displaystyle\implies\theta=(2m+1)\pi+\frac\pi4$

$\displaystyle\implies-\pi<(2m+1)\pi+\frac\pi4<\pi$

$\displaystyle\implies ? <m<?$

Answer 2

Hint : Do you know that $\tan(\theta)$ is +ve in the third quadrant and $\csc(\theta)$ is -ve in the third quadrant ?