Question : If $\sin A=\frac{5}{13}$ and $7 \cot B=24$, then the value of $(\sec A \cos B)(\operatorname{cosec} B \tan A)$ is:
Option 1: $\frac{65}{42}$
Option 2: $\frac{13}{14}$
Option 3: $\frac{15}{13}$
Option 4: $\frac{13}{7}$
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Correct Answer: $\frac{65}{42}$
Solution :
Given that $\sin A=\frac{5}{13}$ and $7 \cot B=24$,
$\sin A=\frac{5}{13}$,
Use the identity $\sin^2 A + \cos^2 A = 1$
$⇒\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}$
$\therefore\tan A = \frac{\sin A}{\cos A} = \frac{5}{12}$
$\therefore\sec A = \frac{1}{\cos A} = \frac{13}{12}$
Given, $7 \cot B=24$
$\therefore\cot B = \frac{24}{7}$
$⇒\operatorname{cosec} B= \sqrt{1 + \cot^2 B} = \sqrt{1 + \left(\frac{24}{7}\right)^2} = \frac{25}{7}$
$⇒\cos B = \sqrt{1-\frac{1}{{\operatorname{cosec}^2 B}} }=\sqrt{1-(\frac{7}{{25}})^2 }=\frac{24}{{25}}$
$⇒(\sec A \cos B)(\operatorname{cosec} B \tan A) = \left(\frac{13}{12} \times \frac{24}{25}\right) \left(\frac{25}{7} \times \frac{5}{12}\right) = \frac{65}{42}$
Hence, the correct answer is $\frac{65}{42}$.
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