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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 |
