If cosθ= frac1sqrt 10 then find the value of sec2θ + tanθ.

cosθ=\(\frac{1}{\sqrt {10}}\)=\(\frac{B}{H}\)

P=\(\sqrt {(\sqrt {10})^2-(1)^2}\)    (H² = B² + P²)

P=\(\sqrt {10-1}\) 

P = 3

Now, sec²θ + tanθ

=(\(\frac{\sqrt {10}}{1}\))²+\(\frac{3}{1}\)

= 10 + 3 = 13