If $cosec θ=\frac{13}{12}$, then the value of $\frac{2sinθ-3cosθ}{4sinθ-9cosθ}$ is:

2 4 1 3

3

cosecθ = \(\frac{13 }{12}\) P² +  B² = H² (12)² + B² = (13)² B = 5 Now , \(\frac{2sinθ - 3cosθ }{ 4sinθ - 9cosθ}\) = \(\frac{2×12 - 3×5 }{ 4×12 - 9×5}\) =  \(\frac{24 - 15 }{3}\) = 3