Correct Answer – Option 1 : 3

Given:

cosec θ = 13/12

Forumula used:

H = hypotenuse, P = perpendicular, B = base

H² = P² + B²

cosecθ = H/P, sinθ = P/H, cosθ = B/P

Calculation:

⇒ cosec θ = 13/12 = H/P

⇒ H = 13, P = 12, B² = H² – P²

⇒ B² = 13² – 12² = 5²

⇒ B = 5

⇒ sinθ = 12/13, cosθ = 5/13, and cosec θ = 13/12

putting the value of sinθ and cosθ in \( \frac{{2sin\theta – 3cos\theta }}{{4sin\theta – 9cos\theta }}\)

⇒ \(\frac{{2 \times \frac{{12}}{{13}} – 3 \times \frac{5}{{13}}}}{{4 \times \frac{{12}}{{13}} – 9 \times \frac{5}{{13}}}}\)

⇒ 3

∴ The value of \( \frac{{2sin\theta – 3cos\theta }}{{4sin\theta – 9cos\theta }}\) is 3.