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

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 = \(\frac{sin θ}{cos θ}\) = \(\frac{\frac{5}{13}}{\frac{12}{13}}\) = \(\frac{5}{12}\)