Question 1

Pythagorean Identities (derived from $\sin^2 A + \cos^2 A = 1$):

$1 + \cot^2 A = \csc^2 A \quad \Rightarrow \quad \csc^2 A - \cot^2 A = 1$

$1 + \tan^2 A = \sec^2 A \quad \Rightarrow \quad \sec^2 A - \tan^2 A = 1$

Reciprocal relationships:

$\sin A = \frac{1}{\csc A}, \quad \tan A = \frac{1}{\cot A}, \quad \sec A = \frac{1}{\cos A}$

We use these to rewrite $\sin A$, $\sec A$, and $\tan A$ purely in terms of $\cot A$.

We will use fundamental trigonometric identities to express the ratios.

Step 1 — Express $\sin A$

Let's start with a basic identity.

We know that $\csc^2 A - \cot^2 A = 1$.

We can rearrange this identity.

$\csc^2 A = 1 + \cot^2 A$

Now, let's take the square root.

$\csc A = \sqrt{1 + \cot^2 A}$

We also know that $\sin A$ is the reciprocal of $\csc A$.

$\sin A = \frac{1}{\csc A}$

Let's substitute the value of $\csc A$.

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

$\boxed{\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}}$

Step 2 — Express $\sec A$

Let's use another important identity.

We know that $\sec^2 A - \tan^2 A = 1$.

We can rearrange this identity.

$\sec^2 A = 1 + \tan^2 A$

We also know that $\tan A$ is the reciprocal of $\cot A$.

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

Let's substitute this into the identity for $\sec^2 A$.

$\sec^2 A = 1 + \left(\frac{1}{\cot A}\right)^2$

Let's simplify the expression.

$\sec^2 A = 1 + \frac{1}{\cot^2 A}$

Let's combine the terms on the right side.

$\sec^2 A = \frac{\cot^2 A + 1}{\cot^2 A}$

Now, let's take the square root.

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

We can simplify the denominator.

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

For acute angles, $\cot A$ is positive.

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

$\boxed{\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}}$

Step 3 — Express $\tan A$

This is a direct reciprocal relationship.

We know that $\tan A$ is the reciprocal of $\cot A$.

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

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

Answer

(i) $\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$ (ii) $\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$ (iii) $\tan A = \frac{1}{\cot A}$