If $\frac{\cos ^2 \theta}{\cot ^2 \theta+\sin ^2 \theta-1}=3, 0^{\circ}<\theta<90^{\circ}$, then the value of $(\tan \theta+{cosec} \theta)$ is:
Answer & explanation
Correct answer: option 1
We are given ,
\(\frac{cos²θ }{cot²θ + sin²θ - 1 }\) = 3
\(\frac{cos²θ }{cot²θ - cos²θ }\) = 3
{ using , sin²θ + cos²θ = 1 }
taking out cos²θ common
\(\frac{1 }{cosec²θ - 1 }\) = 3
1 = 3.cosec²θ - 3
3.cosec²θ = 4
sin²θ = \(\frac{3}{4 }\)
sinθ = \(\frac{√3}{2 }\)
{ we know, sin60º = \(\frac{√3}{2 }\) }
So, θ = 60º
Now,
( tanθ+ cosecθ )
= ( tan60º + cosec60º )
= √3 + \(\frac{2}{√3 }\)
= \(\frac{5}{√3 }\)
= \(\frac{5√3}{3 }\)