If $\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}=4 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\cot \theta+{cosec}~ \theta$ is :

$\sqrt{3}$

We are given that ,

\(\frac{1}{1 - sinθ}\) + \(\frac{1}{1 + sinθ}\) = 4 secθ

\(\frac{2}{1² -  sin²θ}\) = 4 secθ

{ using , sin²θ + cos²θ = 1 }

\(\frac{2}{  cos²θ}\) = 4 secθ

sec²θ = 2 secθ

secθ = 2

{ we know, sec60º = 2 }

So, θ = 60º

Now,

cotθ + cosecθ

= cot60º + cosec60º

= \(\frac{1}{ √3 }\) + \(\frac{2}{ √3 }\)

= \(\frac{3}{ √3 }\)

= √3