If sin \(\theta\) = \(\frac{m - n}{m + n}\); Find the value of 1 + tan2\(\theta\)

A.

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

B.

\(\frac{(m^2 + n^2 + 2mn)}{4mn}\)

C.

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

D.

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

Correct answer is B

\((m + n)^{2} = (m - n)^{2} + x^{2}\)

\(m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}\)

\(x^{2} = 4mn\)

\(x = \sqrt{4mn} = 2\sqrt{mn}\)

1 + tan2\(\theta\) = sec2\(\theta\)

= \(\frac{1}{cos^2\theta}\)

\(\cos \theta = \frac{2\sqrt{mn}}{(m + n)}\)

\(\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}\)

\(\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}\)

= \(\frac{(m^2 + n^2 + 2mn)}{4mn}\)