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If $12 \text{cosec } A = 13$, then find the value of $\frac{2 \text{sin } A - 3 \text{cos } A}{4 \text{sin } A - 9 \text{cos } A}$.
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Sol. $\text{sin } A = \frac{12}{13}$, $\text{cos } A = \frac{5}{13}$
Hence $\frac{2 \text{sin } A - 3 \text{cos } A}{4 \text{sin } A - 9 \text{cos } A} = \frac{2 \times \frac{12}{13} - 3 \times \frac{5}{13}}{4 \times \frac{12}{13} - 9 \times \frac{5}{13}} = 3$
Hence $\frac{2 \text{sin } A - 3 \text{cos } A}{4 \text{sin } A - 9 \text{cos } A} = \frac{2 \times \frac{12}{13} - 3 \times \frac{5}{13}}{4 \times \frac{12}{13} - 9 \times \frac{5}{13}} = 3$