If cosθ=\(\frac{1}{\sqrt {26}}\), then find the value of sec2θ + tanθ.

18 5 31 27

31

cosθ = \(\frac{1}{\sqrt {26}}\)=\(\frac{B}{H}\) P = \(\sqrt {(\sqrt {26})^2-(1)^2}\)    (H2 = B2 + P2) P = \(\sqrt {26-1}\)  P = 5 Now, sec2θ + tanθ = (\(\frac{\sqrt {26}}{1}\))2 + \(\frac{5}{1}\) = 26 + 5 = 31