Question

If $\sin \theta =\frac{m^2-n^2}{m^2+n^2}$, determine the values of $\tan \theta$, $\sec \theta$ and $\csc \theta$.

Answer 1

We know that
${(m^2+n^2)}^2-{(m^2+n^2)}^2=4m^2{n}^2$.
$\therefore \tan \theta \frac{m^2-n^2}{2mn}$, $\sec \theta =\frac{m^2+n^2}{2mn}$,
$\csc \theta =\frac{m^2+n^2}{m^2-n^2}$
You have to make a triangle whose height is $m^2-n^2$ and hypotenuse is $m^2-n^2$ so that its base is $2mn$ etc. as explained in $\S 2\left(6\right),P.508$