If $\frac{cosec~\theta+\cot \theta}{cosec~\theta-\cot \theta}=7$, then the value of $\frac{4 \sin ^2 \theta-1}{4 \sin ^2 \theta+5}$ is:

$\frac{1}{9}$

We are given that :-

\(\frac{ cosecθ + cotθ}{cosecθ - cotθ }\) = 7

cosecθ + cotθ = 7cosecθ - 7cotθ

8 cotθ = 6 cosecθ

cosθ = \(\frac{ 3}{4}\)

Now, cos²θ = \(\frac{ 9}{16}\)

{ using , sin²θ  + cos²θ  = 1 }

1 - sin²θ  = \(\frac{ 9}{16}\)

sin²θ = 1 - \(\frac{ 9}{16}\)

sin²θ = \(\frac{ 7}{16}\)

Now,

\(\frac{ 4sin²θ  - 1 }{4sin²θ  + 5 }\)

= \(\frac{ 4 × 7/16 - 1 }{4×7/16  + 5 }\)

= \(\frac{ 3/4 }{27/4 }\)

= \(\frac{ 1 }{9 }\)