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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