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

-14

\(\frac{sin²θ}{tan²θ - sin²θ }\) = 5

\(\frac{sin²θ}{sin²θ/cos²θ - sin²θ }\) = 5

\(\frac{sin²θ}{sin²θ.sec²θ - sin²θ }\) = 5

\(\frac{sin²θ}{sin²θ(sec²θ -1) }\) = 5

{ using , sec²θ -  tan²θ = 1 }

\(\frac{sin²θ}{sin²θ(tan²θ) }\) = 5

cot²θ = 5

cotθ =  \(\frac{√5 }{1 }\)

{ We know, cotθ =  \(\frac{B }{P}\) }

By using pythagoras theorem ,

P² + B² = H²

1² + 5 = H²

H = √6

Now,

\(\frac{24sin²θ - 15sec²θ}{6cosec²θ - 7cot²θ }\)

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

= \(\frac{4 - 18}{36 - 35 }\)

= -14