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

1

We are given,

cot θ = \(\frac{1}{ √3 }\)

{ we know, cot 60º = \(\frac{1}{ √3 }\) }

So, θ = 60º

Now,

\(\frac{2 - sin² θ}{ 1 - cos² θ }\) + ( cosec² θ - sec θ )

= \(\frac{2 - sin² 60º}{ 1 - cos² 60º }\) + ( cosec² 60º - sec 60º )

= \(\frac{2 - 3/4}{ 1 - 1/4 }\) + ( 4/3 - 2 )

= \(\frac{5/4}{ 3/4 }\) + ( 4/3 - 2 )

= \(\frac{5}{ 3}\) + (-2/3 )

= 1