If $\frac{cosecθ+cotθ}{cosecθ-cotθ}=7,$ then the value of $\frac{4sin^2θ+5}{4sin^2θ-1}$ is :

9

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

\(\frac{1+ cosθ}{1 - cosθ}\) = 7

1+ cosθ = 7 - 7cosθ

8cosθ = 6

cosθ = \(\frac{3}{4}\)

{ we know, cosθ = \(\frac{B}{H}\) }

By using pythagoras theorem,

P² + B² = H²

P² + 3² = 4²

P = \(\sqrt {7 }\)

Now,

\(\frac{4 sin²θ + 5 }{4sin²θ - 1 }\)

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

= 9