If 5sinθ =  12cosθ, then what is the value of secθ ?

\(\frac{13}{5}\)  \(\frac{13}{12}\) 1  \(\frac{5}{13}\)

\(\frac{13}{5}\)

5sinθ =  12cos θ ⇒ \(\frac{sin θ}{cos θ}\) = \(\frac{12}{5}\) ⇒ tan θ = \(\frac{12 (P)}{5 (B)}\) using Pythagoras theorem ,  Hypotenuse = 13 secθ = \(\frac{H}{B}\)= \(\frac{13}{5}\)