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:

$\frac{4+\sqrt{2}}{2}$

\(\frac{1}{cosecθ + 1 }\) + \(\frac{1}{cosecθ - 1 }\) = 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,

\(\frac{tanθ + 2secθ}{cosecθ}\)

= \(\frac{tan45º + 2sec45º}{cosec45º}\)

= \(\frac{1 + 2√2}{ √2 }\)

= \(\frac{4+ √2}{ 2 }\)