If $3 \sin ^2 \theta+4 \cos \theta-4=0,0^{\circ}<\theta<90^{\circ}$, then the value of $\left(cosec^2 \theta+\cot ^2 \theta\right)$ is

$\frac{5}{4}$

3 sin²θ + 4 cosθ - 4 = 0

3 ( 1 - cos²θ) + 4 cosθ - 4 = 0

3cos²θ - 4 cosθ + 1 = 0

on solving ,

cosθ = \(\frac{1}{3}\)

By using pythagoras theorem ,

P² + B² = H²

P² + 1² = 3²

P = 2√2

Now , ( cosec²θ + cot²θ)

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

= \(\frac{9}{8}\) + \(\frac{1}{8}\)

= \(\frac{5}{4}\)