Question

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

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Answer

To solve the problem where $$\csc A = \frac{13}{12}$$, and we need to find $$\frac{2 \sin A - 3 \cos A}{4 \sin A - 9 \cos A},$$ follow these steps:

  1. Identify $\sin A$ and $\cos A$:

    • Since $\csc A = \frac{1}{\sin A}$, we have $$\sin A = \frac{12}{13}.$$

  2. Calculate $\cos A$:

    • Using the Pythagorean identity, $$ \sin^2 A + \cos^2 A = 1, $$ we substitute $\sin A$: $$ \left(\frac{12}{13}\right)^2 + \cos^2 A = 1. $$ Solving for $\cos^2 A$ gives: $$ \cos^2 A = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}. $$ Thus, $$ \cos A = \sqrt{\frac{25}{169}} = \frac{5}{13}, $$ considering positive root, as $\cos A$ corresponds to an angle in a typical right triangle context here (assuming $0 \leq A \leq \frac{\pi}{2}$).

  3. Simplify the given expression:

    • Substitute the values of $\sin A$ and $\cos A$ into the expression: $$ \frac{2 \sin A - 3 \cos A}{4 \sin A - 9 \cos A} = \frac{2 \cdot \frac{12}{13} - 3 \cdot \frac{5}{13}}{4 \cdot \frac{12}{13} - 9 \cdot \frac{5}{13}}. $$

    • Simplifying the expression: $$ \frac{24 - 15}{48 - 45} \rightarrow \frac{9}{3} = 3. $$

Conclusion: The value of $$\frac{2 \sin A - 3 \cos A}{4 \sin A - 9 \cos A}$$ is 3.


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