If cosθ=\(\frac{1}{\sqrt {17}}\) then find the value of sec2θ + tanθ.

15 17 18 21

21

cosθ=\(\frac{1}{\sqrt {17}}\)=\(\frac{B}{H}\) P=\(\sqrt {(\sqrt {17})^2-(1)^2}\)    (H2 = B2 + P2) P=\(\sqrt {17-1}\)  P = 4 Now, sec2θ + tanθ =(\(\frac{\sqrt {17}}{1}\))2+\(\frac{4}{1}\) = 17 + 4 = 21