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

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=\(\frac{1}{2}\)+\(\frac{5}{4}\)

= \(\frac{7}{4}\)