Originally Posted by
Soroban Hello, tempq1!
$\displaystyle \text}We are given: }\;\cos\theta \:=\:\dfrac{m^2-n^2}{m^2+n^2} \;=\;\dfrac{adj}{hyp}$
$\displaystyle \cs\theta$ is in a right triangle with: .$\displaystyle adj \,=\, m^2-n^2,\;hyp \,=\, m^2+n^2$
Using Pythagorus, we find that: .$\displaystyle opp \,=\,\pm 2mn$
And we can now express $\displaystyle \sin\theta$ and $\displaystyle \tan\theta.$
To apply the correct signs to our answer, we note that:
. . $\displaystyle \cos\theta$ is positive: $\displaystyle \cs\theta$ is in Quadrant 1 or 4.
. . $\displaystyle \csc\theta$ is negative: $\displaystyle \cs\theta$ is in Quadrant 3 or 4.
. . . . Hence, $\displaystyle \cs\theta$ is in Quadrant 4.
Go for it!