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

$$6tan\theta-5\sqrt3sec\theta+12cot\theta=0,0^{0}<\theta<90^{0}$$

$$\frac{6sin\theta}{cos\theta}-5\sqrt3(\frac{1}{cos\theta}) + \frac{12cos\theta}{sin\theta} = 0

$$\frac{6sin^2\theta - 5\sqrt3sin\theta + 12cos^2\theta}{sin\theta cos\theta}$$ = 0

$$\frac{6sin^2\theta + 6cos^2\theta- 5\sqrt3sin\theta + 6cos^2\theta}{sin\theta cos\theta}$$ = 0

$$\frac{6- 5\sqrt3sin\theta + 6cos^2\theta}{sin\theta cos\theta}$$ = 0

$$6- 5\sqrt3sin\theta + 6(1 - sin^2 \theta) = 0$$

$$6sin^2 \theta +  5\sqrt3sin\theta - 12 = 0$$

From LHS,

Put the $$\theta = 60 \degree$$

$$6sin^2 \theta + 5\sqrt3sin\theta - 12$$

$$6sin^2 60 \degree + 5\sqrt3sin60 \degree - 12$$

$$6(\frac{\sqrt3}{2})^2 + 5\sqrt3 \times \frac{\sqrt3}{2} - 12$$

$$\frac{18}{4} + \frac{15}{2} - 12$$ = 0

= RHS

$$(cosec\theta+sec\theta)$$

Put the $$\theta = 60 \degree$$,

= $$(cosec60 \degree+sec60 \degree)$$

= $$\frac{2}{\sqrt3} + 2$$

= $$\frac{2 + 2\sqrt3}{\sqrt3}$$

= $$\frac{2\sqrt3 + 6}{3}$$

= $$\frac{2}{3}(\sqrt3 + 3)$$