If 2\(\sqrt {5}\) sinθ = 2 find the value of (tanθ + sec2θ).

\(\frac{3}{4}\) \(\frac{6}{4}\) \(\frac{7}{4}\) \(\frac{3}{5}\)

\(\frac{7}{4}\)

2\(\sqrt {5}\) sinθ = 2 sinθ = \(\frac{2}{2\sqrt {5}}\)=\(\frac{P}{H}\) B=\(\sqrt {(H)^2-P^2}\) B=\(\sqrt {(2\sqrt {5})^2-(2)^2}\) B = 4 Now, (tanθ + sec2θ) \(\frac{2}{4}\)+(\(\frac{2\sqrt {5}}{4}\))2=\(\frac{1}{2}\)+\(\frac{5}{4}\) = \(\frac{7}{4}\)