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