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

Using trigonometric formulas,

$\cos e{c}^{2}A – co{t}^{2}A = 1$

$\cos e{c}^{2}A = 1 + co{t}^{2}A$

Using $\cos ec A = \frac{1}{\sin A}$.

$\Rightarrow \frac{1}{{\sin }^{2}A} = 1 + co{t}^{2}A$

Rearranging further we get,

$\Rightarrow {\sin }^{2}A = \frac{1}{1 + co{t}^{2}A}$

Now, take square roots on both sides, we get

$\Rightarrow \sin A = \pm \frac{1}{\sqrt{1 + co{t}^{2}A}}$

Hence we get the required result for sine function.

Now using the result,

${\sin }^{2}A = \frac{1}{1 + co{t}^{2}A}$

As we know, $\Rightarrow {\sin }^{2}A = 1 – {\cos }^{2}A$

$\Rightarrow 1 – {\cos }^{2}A = \frac{1}{1 + co{t}^{2}A}$

Rearranging further we get,

${\cos }^{2}A =1- \frac{1}{1 + co{t}^{2}A}$

$\Rightarrow {\cos }^{2}A = \frac{1 + co{t}^{2}A-1}{1 + co{t}^{2}A}=\frac{ co{t}^{2}A}{1 + co{t}^{2}A}$

Using, $Cos A = \frac{1}{sec A}$

$\Rightarrow \frac{ 1}{se{c}^{2}A} =\frac{ co{t}^{2}A}{1 + co{t}^{2}A}$

$\Rightarrow se{c}^{2}A =\frac{1+ co{t}^{2}A}{ co{t}^{2}A}$

Now, square roots on both sides, we get

$secA =\pm \sqrt{\frac{1+ co{t}^{2}A}{ co{t}^{2}A}}=\pm \frac{\sqrt{1+ co{t}^{2}A}}{cot A}$

Hence we get the required result for secant function.

Now, as we know $\tan A=\frac{1}{cot A}$

Hence we get the required result for tangent function.