If $\frac{\cos ^2 \theta}{\cot ^2 \theta+\sin ^2 \theta-1}=3, 0^{\circ}<\theta<90^{\circ}$, then the value of $(\tan \theta+{cosec} \theta)$ is:

$\frac{5 \sqrt{3}}{3}$

We are given ,

\(\frac{cos²θ }{cot²θ + sin²θ  - 1 }\) = 3

\(\frac{cos²θ }{cot²θ - cos²θ   }\) = 3

{ using , sin²θ + cos²θ = 1 }

taking out cos²θ  common

\(\frac{1 }{cosec²θ - 1   }\) = 3

1 = 3.cosec²θ - 3

3.cosec²θ = 4

sin²θ = \(\frac{3}{4 }\)

sinθ = \(\frac{√3}{2 }\)

{ we know, sin60º = \(\frac{√3}{2 }\) }

So, θ = 60º

Now,

( tanθ+ cosecθ )

= ( tan60º + cosec60º ) 

= √3 + \(\frac{2}{√3 }\)

= \(\frac{5}{√3 }\)

= \(\frac{5√3}{3 }\)