\(\frac{\sqrt{n^2 - m^2}}{m}\)
\(\frac{\sqrt{(n + m)(n - m)}}{n}\)
\(\frac{m}{\sqrt{n^2 - m^2}}\)
\(\sqrt{\frac{n}{n^2 - m^2}}\)
Correct answer is B
sin \(\theta\) = \(\frac{m}{n}\)
Opp = m; Hyp = n
Adj = \(\sqrt{n^{2} - m^{2}}\)
\(\cos \theta = \frac{\sqrt{n^{2} - m^{2}}}{n}\)
= \(\frac{\sqrt{(n + m)(n - m)}}{n}\)