Question

Write all the other trigonometric ratios of $\angle A$ in terms of $\text{sec}A.$

Answer 1

We know that,

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

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

Also,

${\text{cos}}^{2}A+{\text{sin}}^{2}A=1$

$\Rightarrow {\text{sin}}^{2}A=1-{\text{cos}}^{2}A$

$\Rightarrow {\text{sin}}^{2}A=1-\frac{1}{\left({\text{sec}}^{2}A\right)}$

$\Rightarrow {\text{sin}}^{2}A=\frac{({\text{sec}}^{2}A-1)}{{\text{sec}}^{2}A}$

$\Rightarrow \text{sin}A=\pm \frac{\sqrt{{\text{sec}}^{2}A-1}}{\text{sec}A}\dots \dots \left(i\right)$

$\text{sin}A=\frac{1}{\text{cosec}A}$

$\Rightarrow \text{cosec}A=\frac{1}{\text{sin}A}$

$\Rightarrow \text{cosec}A=\pm \frac{\text{sec}A}{\sqrt{{\text{sec}}^{2}A-1}}$ [Using $eq.\left(i\right)$]

${\text{sec}}^{2}A-{\text{tan}}^{2}A=1$

$\Rightarrow {\text{tan}}^{2}A={\text{sec}}^{2}A-1$

$\Rightarrow \text{tan}A=\sqrt{{\text{sec}}^{2}A-1}$

$\text{tan}A=\frac{1}{\text{cot}A}$

$\Rightarrow \text{cot}A=\frac{1}{\text{tan}A}$

$\Rightarrow \text{cot}A=\pm \frac{1}{\sqrt{{\text{sec}}^{2}A-1}}$