Question : If $\sec A=\frac{9}{4}$, then what is the value of $\cot A$?
Option 1: $\frac{4}{\sqrt{65}}$
Option 2: $\frac{9}{\sqrt{65}}$
Option 3: $\frac{\sqrt{65}}{9}$
Option 4: $\frac{\sqrt{65}}{4}$
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Correct Answer: $\frac{4}{\sqrt{65}}$
Solution : Given: $\sec A=\frac{9}{4}$ We know that, $\sec^2 A-\tan^2 A=1$ ⇒ $\tan^2 A=\sec^2 A-1$ Putting the value, we get: ⇒ $\tan^2 A=(\frac{9}{4})^2-1$ ⇒ $\tan^2 A=\frac{81}{16}-1$ ⇒ $\tan^2 A=\frac{65}{16}$ ⇒ $\tan A=\frac{\sqrt{65}}{4}$ $\therefore \cot A=\frac{4}{\sqrt{65}}$ Hence, the correct answer is $\frac{4}{\sqrt{65}}$.
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