If $\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}=4 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\cot \theta+{cosec}~ \theta$ is :
Answer & explanation
Correct answer: option 2
We are given that ,
\(\frac{1}{1 - sinθ}\) + \(\frac{1}{1 + sinθ}\) = 4 secθ
\(\frac{2}{1² - sin²θ}\) = 4 secθ
{ using , sin²θ + cos²θ = 1 }
\(\frac{2}{ cos²θ}\) = 4 secθ
sec²θ = 2 secθ
secθ = 2
{ we know, sec60º = 2 }
So, θ = 60º
Now,
cotθ + cosecθ
= cot60º + cosec60º
= \(\frac{1}{ √3 }\) + \(\frac{2}{ √3 }\)
= \(\frac{3}{ √3 }\)
= √3