If cos\(\theta\) = \(\frac{a}{b}\), find 1 + tan2\(\theta\)

A.

\(\frac{b^2}{a^2}\)

B.

\(\frac{a^2}{b^2}\)

C.

\(\frac{a^2 + b^2}{b^2 - a^2}\)

D.

\(\frac{2a^2 + b^2}{a^2 + b^2}\)

Correct answer is A

cos\(\theta\) = \(\frac{a}{b}\), Sin\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a}}\)

Tan\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a^2}}\), Tan 2 = \(\sqrt{\frac{b^2 - a^2}{a^2}}\)

1 + tan2\(\theta\) = 1 + \(\frac{b^2 - a^2}{a^2}\)

= \(\frac{a^2 + b^2 - a^2}{a^2}\)

= \(\frac{b^2}{a^2}\)