If 5sin θ + 12cos θ = 13 ; then find value of tan θ

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

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

5sin θ + 12cos θ = 13 ⇒ \(\frac{5}{13}\)sin θ + \(\frac{12}{13}\)cos θ = 1 .....(i) *(sin2θ + cos2θ = 1)* Therefore, from (i) sin θ = \(\frac{5}{13}\) & cos θ = \(\frac{12}{13}\) ⇒  tan θ = \(\frac{sin θ}{cos θ}\) = \(\frac{\frac{5}{13}}{\frac{12}{13}}\) = \(\frac{5}{12}\)