If \(\cos \theta =\frac{m^{2}-n^{2}}{m^{2}+n^{2}}\), then what is \(\tan \theta \) ?
- AOption (a)
\(\frac{2mn}{m^{2}-n^{2}} \)
- BOption (b)
\(\frac{m^{2}-n^{2}}{2mn} \)
- COption (c)
\(\frac{m^{2}+n^{2}}{2mn}\)
- DOption (d)
\(\frac{2mn}{m^{2}+n^{2}}\)
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Correct answer
\(\frac{2mn}{m^{2}-n^{2}} \)