The value of $\frac{sec θ \, cosec θ}{2+tan^2θ+cot^2θ}$ is equal to :
Answer & explanation
Correct answer: option 3
\(\frac{sec θ.cosec θ}{2 + tan²θ+cot² θ }\)
= \(\frac{sec θ.cosec θ}{1 + tan²θ +1 +cot² θ }\)
{ sec²θ - tan²θ = 1 & cosec²θ - cot²θ = 1 }
= \(\frac{sec θ.cosec θ}{ sec²θ . cosec² θ }\)
= \(\frac{1}{ secθ . cosec θ }\)
= cosθ . sin θ