If tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\) find sec\(\theta\)

A.

\(\frac{m^2 + n^2}{(m^2 - n^2)}\)

B.

\(\frac{m^2 + n^2}{2mn}\)

C.

\(\frac{mn}{2(m^2 + n^2)}\)

D.

\(\frac{m^2n^2}{2(m^2 - n^2)}\)

Correct answer is B

Tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\)

\(\frac{\text{Opp}}{\text{Adj}}\) by pathagoras theorem

= Hyp2 = Opp2 + Adj2

Hyp2 = (m2 - n2) + (2mn)2

Hyp2 = m4 - 2m2n4 - 4m2 - n2

Hyp2 = m4 + 2m2 + n2n

Hyp2 = (m2 - n2)2

Hyp2 = \(\frac{m^2 + n^2}{2mn}\)