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

$\frac{19}{12}$

tan θ + 3cot θ - 2√3 = 0

tan θ + 3 × \(\frac{1}{tan θ}\) - 2√3 = 0

tan² θ - 2√3tan θ  +  3 = 0

tan² θ - √3tan θ - √3tan θ  +  3 = 0

tan θ ( tan θ - √3 ) - 1 ( tan θ - √3 ) = 0

( tan θ - 1 ). ( tan θ - √3 ) = 0

Either ( tan θ - 1 ) = 0 Or  ( tan θ - √3 ) = 0 

tan θ - 1 = 0 is not possible because  0º < θ < 90º

So, tan θ - √3 = 0

tan θ = √3

{ we know, tan60º = √3 }

Now,

( cosec² θ + cot² θ )

= ( cosec² 60º + cos² 60º ) 

= \(\frac{4}{3}\) + \(\frac{1}{4}\)

= \(\frac{19}{12}\)