If $\frac{1}{cosec \theta + 1} + \frac{1}{cosec \theta -1} = 2 \sec \theta, 0^\circ < \theta < 90^\circ$, then the value of $\frac{\tan \theta + 2 \sec \theta}{cosec \theta}$ is:
Answer & explanation
Correct answer: option 3
\(\frac{1}{cosecθ + 1 }\) + \(\frac{1}{cosecθ - 1 }\) = 2secθ
\(\frac{cosecθ - 1 + cosecθ + 1}{cosec²θ - 1² }\) = 2secθ
{ using , cosec²θ - 1² = cot²θ }
2cosecθ = 2secθ
cotθ = 1
{ we know, cot45º = 1 }
So, θ = 45º
Now,
\(\frac{tanθ + 2secθ}{cosecθ}\)
= \(\frac{tan45º + 2sec45º}{cosec45º}\)
= \(\frac{1 + 2√2}{ √2 }\)
= \(\frac{4+ √2}{ 2 }\)