If cosθ=\(\frac{1}{\sqrt {26}}\),
then find the value of sec2θ + tanθ.
Answer & explanation
Correct answer: option 3
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