Express the trigonometric ratios $\sin A$,$\sec A$,and $\tan A$ in terms of $\cot A$.

(N/A) We know that,$\operatorname{cosec}^{2} A = 1 + \cot^{2} A$.
Since $\sin A = \frac{1}{\operatorname{cosec} A}$,we have $\sin^{2} A = \frac{1}{\operatorname{cosec}^{2} A} = \frac{1}{1 + \cot^{2} A}$.
Taking the square root,$\sin A = \frac{1}{\sqrt{1 + \cot^{2} A}}$.
Next,we know that $\tan A = \frac{1}{\cot A}$.
Finally,using the identity $\sec^{2} A = 1 + \tan^{2} A$,we substitute $\tan A = \frac{1}{\cot A}$:
$\sec^{2} A = 1 + \left(\frac{1}{\cot A}\right)^{2} = 1 + \frac{1}{\cot^{2} A} = \frac{\cot^{2} A + 1}{\cot^{2} A}$.
Taking the square root,$\sec A = \frac{\sqrt{1 + \cot^{2} A}}{\cot A}$.