Question

If $\sin \theta =\frac{m^2-n^2}{m^2+n^2}$ then $\tan \theta$.

Answer 1

Let,

$\sin \theta =\frac{m^2-n^2}{m^2+n^2}$

or, $cosec\theta =\frac{m^2+n^2}{m^2-n^2}$.

We have,

$\cot \theta =\sqrt{cose{c}^2\theta -1}$

or, $\cot \theta =\sqrt{{\left(\frac{m^2+n^2}{m^2-n^2}\right)}^2-1}$

or, $\cot \theta =\sqrt{{\left(\frac{2mn}{m^2-n^2}\right)}^2}$

or, $\tan \theta =\frac{m^2-n^2}{2mn}$