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