Write all the other trigonometric ratios of ∠A in terms of sec A.

As we know,

$sec A = \frac{1}{\cos A}$

$\Rightarrow \cos A =\frac{1}{sec A}$

So we get Cos A function in terms of sec A.

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

using $\cos A =\frac{1}{sec A}$

$\Rightarrow {\sin }^{2}A = 1 –\frac{1}{se{c}^{2}A}$

$\Rightarrow {\sin }^{2}A = \frac{se{c}^{2}A-1}{se{c}^{2}A}$

Take square root on both sides,

$\sin A =\pm \sqrt{\frac{se{c}^{2}A-1}{se{c}^{2}A}} =\pm \frac{\sqrt{se{c}^{2}A-1}}{sec A}$

So we get Sin A function in terms of sec A.

As we know, $\cos ec A=\frac{1}{\sin A}$

using $\sin A =\pm \frac{\sqrt{se{c}^{2}A-1}}{sec A}$

$\cos ecA =\pm \frac{sec A}{\sqrt{se{c}^{2}A-1}}$

So we get cosec A function in terms of sec A.

As we know

$se{c}^{2}A – {\tan }^{2}A = 1$

Rearrange the terms,

$\Rightarrow {\tan }^{2}A = se{c}^{2}A – 1$

Take square root on both sides,

$\tan A = \pm \sqrt{se{c}^{2}A – 1}$

So we get tan A function in terms of sec A.

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

Using $\tan A = \pm \sqrt{se{c}^{2}A – 1}$

$\Rightarrow cotA =\pm \frac{1}{ \sqrt{se{c}^{2}A – 1}}$

So we get cot A function in terms of sec A.