If 5 cos T + 12 sin T = 13, then tan T = ;

5 cosθ + 12 sinθ = 13

Dividing by cosθ, 5 + 12 tanθ = 13 secθ

on squaring, 25 + 144 tan2θ + 120 tanθ

= 169 sec2θ = 169 (1 + tan2θ)

⇒ 169 tan2θ – 144 tan2θ – 120 tan θ = 169 – 25

⇒ 25 tan2θ – 120 tan θ + 144 = 0

⇒ (5 tan θ – 12)2 = 0

⇒ 5 tan θ = 12

$Rightarrow an heta=frac{12}{5}$