Question

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

Answer 1

Step 1: Express $\mathbf{sin}\mathbf{}\mathit{A}$ in terms of $\mathbf{cot}\mathbf{}\mathit{A}$.

It is known that, ${\mathrm{cosec}}^{2}A=1+{\cot }^{2}A$.

$$\begin{gathered} \Rightarrow \mathrm{cosec}A=\sqrt{1+{\cot }^{2}A} \\ \Rightarrow \frac{1}{\mathrm{cosec}A}=\frac{1}{\sqrt{1+{\cot }^{2}A}} \end{gathered}$$

It is also known that, $\sin A=\frac{1}{\mathrm{cosec}A}$.

Thus, $\sin A=\frac{1}{\sqrt{1+{\cot }^{2}A}}$.

Step 2: Express $\mathbf{sec}\mathbf{}\mathit{A}$ in terms of $\mathbf{cot}\mathbf{}\mathit{A}$.

It is known that, ${\sec }^{2}A=1+{\tan }^{2}A$.

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

It is also known that, $\tan A=\frac{1}{\cot A}$.

Thus, $\sec A=\sqrt{1+\frac{1}{{\cot }^{2}A}}$.

Step 3: Express $\mathbf{tan}\mathbf{}\mathit{A}$ in terms of $\mathbf{cot}\mathbf{}\mathit{A}$.

It is known that, $\tan A=\frac{1}{\cot A}$.

Thus, $\tan A=\frac{1}{\cot A}$.

Therefore, $\mathbf{sin}\mathbf{}\mathit{A}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{cot}}^{\mathbf{2}}\mathbf{}\mathbf{A}}}$, $\mathbf{sec}\mathbf{}\mathit{A}\mathbf{=}\sqrt{\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{cot}}^{\mathbf{2}}\mathbf{}\mathbf{A}}}$ and $\mathbf{tan}\mathbf{}\mathit{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{cot}\mathbf{}\mathbf{A}}$.