If $6tan\theta-5\sqrt3sec\theta+12cot\theta=0,0^{0}<\theta<90^{0}$, hen the value of $(cosec\theta+sec\theta)$ is:

$\frac{3+2\sqrt{3}}{2}$ $\frac{2(3+2\sqrt{3})}{3}$ $\frac{2}{3}(3+\sqrt{3})$ $\frac{3+\sqrt{3}}{2}$

$\frac{2}{3}(3+\sqrt{3})$

6tanθ -5√3 secθ + 12 cotθ = 0 Let us put θ = 60º 6tan60º -5√3 sec60º + 12 cot60º = 0 6 × √3  -5√3 × 2 + 12 × \(\frac{1}{√3}\) = 0 0 = 0 ( satisfied ) Now, cosecθ + cotθ = cosec60º + cot60º =  \(\frac{2}{√3}\) + 2 = \(\frac{2}{3}\) × ( 3 + √3 )