If cosθ=\(\frac{1}{\sqrt {17}}\)
then find the value of sec2θ + tanθ.
Answer & explanation
Correct answer: option 4
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