If $\frac{sin^2θ}{tan^2θ-sin^2θ}=5,$ then the value of $\frac{24cos^2θ-15sec^2θ}{6cosec^2θ-7cot^2θ}$ is :

2

$\frac{sin^2θ}{tan^2θ-sin^2θ}=5,$

\(\frac{sin²θ}{tan²θ - sin²θ}\) = 5

\(\frac{cos²θ}{1 - cos²θ}\) = 5

cos²θ = 5 - 5cos²θ

6cos²θ = 5

cosθ = \(\frac{√5}{√6}\) 

{ cosθ = \(\frac{B}{H}\)  }

By using pythagoras theorem,

P² + B² = H²

P² + 5 = 6

P = 1

Now, 

\(\frac{24cos²θ - 15sec²θ}{6cosec²θ - 7cot²θ}\)

= \(\frac{24 ×5/6  - 15×6/5}{6×6/1 - 7×5/1}\)

= \(\frac{20  - 18}{36 - 35}\)

= 2