If $5 \sin ^2 \theta-4 \cos \theta-4=0,0^{\circ}<\theta<90^{\circ}$, then the value of $(\cot \theta+{cosec} \theta)$ is:

$\frac{\sqrt{6}}{2}$

We are given that :-

5sin²θ - 4cosθ - 4  = 0 

{ using , sin²θ + cos²θ = 1 }

5( 1 - cos²θ) - 4cosθ - 4  = 0 

5 cos²θ +  4cosθ - 1  = 0 

5 cos²θ +  5cosθ - cosθ - 1  = 0 

5 cosθ (cosθ  + 1 )  - 1 ( cosθ + 1 )  = 0 

Either (5 cosθ - 1 ) = 0 or ( cosθ + 1 ) = 0

(cosθ + 1  ) = 0 is not possible

So, (5 cosθ - 1 ) = 0 

cosθ = \(\frac{1}{5}\)

By using pythagoras theorem,

P² + B² = H²

P² + 1² = 5²

P² = 24

P = 2√6

Now,

( cotθ + cosecθ )

= \(\frac{B}{P}\) + \(\frac{H}{P}\)

= \(\frac{1}{2√6}\) + \(\frac{5}{2√6}\)

= \(\frac{3}{√6}\)

= \(\frac{3}{√6}\) × \(\frac{√6}{√6}\)

= \(\frac{√6}{2}\)