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:
Answer & explanation
Correct answer: option 4
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