If $3 \sin ^2 \theta-\cos \theta-1=0,0^{\circ}<\theta<90^{\circ}$, then what is the value of $\cot \theta+{cosec} \theta$ ?

$\sqrt{5}$

We are given that ,

3 sin²θ - cosθ - 1 = 0

{ using , sin²θ + cos²θ = 1 }

3 ( 1 - cos²θ ) - cosθ - 1 = 0

3 cos²θ + cosθ - 2 = 0

3 cos²θ + 3 cosθ - 2cosθ - 2 = 0

3 cosθ  (cosθ + 1 ) - 2 (cosθ + 1 ) = 0

(3cosθ - 2 ) . (cosθ + 1 ) = 0

Either (3cosθ - 2 ) = 0 or  (cosθ + 1 ) = 0 

(cosθ + 1 ) = 0 is not possible

So, (3cosθ - 2 ) = 0 

cosθ = \(\frac{2}{3}\)

{ we know, cosθ = \(\frac{B}{H}\) }

By using pythagoras theorem,

P² + B² = H²

P² + 2² = 3²

P = √5

Now,

cotθ+ cosecθ

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

= \(\frac{2}{√5}\) + \(\frac{3}{√5}\)

= \(\frac{5}{√5}\)

= √5