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:
Answer & explanation
Correct answer: option 3
\(\frac{sin²θ}{tan²θ - sin²θ }\) = 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