Question 2

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

$\sin^2 A + \cos^2 A = 1$

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

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

Reciprocal relationships:

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

We use these to express every other ratio purely in terms of $\sec A$.

We will use fundamental trigonometric identities to express each ratio.

Step 1 — Express $\cos A$ in terms of $\sec A$

We know the reciprocal identity. $\cos A$ is the reciprocal of $\sec A$.

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

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

Step 2 — Express $\sin A$ in terms of $\sec A$

Let's use the Pythagorean identity. We know $\sin^2 A + \cos^2 A = 1$.

$\sin^2 A = 1 - \cos^2 A$

Now, substitute the value of $\cos A$.

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

$\sin^2 A = 1 - \frac{1}{\sec^2 A}$

Let's find a common denominator.

$\sin^2 A = \frac{\sec^2 A - 1}{\sec^2 A}$

Now, take the square root on both sides.

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

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

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

$\boxed{\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}}$

Step 3 — Express $\tan A$ in terms of $\sec A$

We use another Pythagorean identity. We know $\tan^2 A + 1 = \sec^2 A$.

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

Now, take the square root on both sides.

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

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

Step 4 — Express $\operatorname{cosec} A$ in terms of $\sec A$

We know that $\operatorname{cosec} A$ is the reciprocal of $\sin A$. Let's use the expression for $\sin A$.

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

$\operatorname{cosec} A = \frac{1}{\frac{\sqrt{\sec^2 A - 1}}{\sec A}}$

$\operatorname{cosec} A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$

$\boxed{\operatorname{cosec} A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}}$

Step 5 — Express $\cot A$ in terms of $\sec A$

We know that $\cot A$ is the reciprocal of $\tan A$. Let's use the expression for $\tan A$.

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

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

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

Answer

(i) $\cos A = \frac{1}{\sec A}$ (ii) $\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$ (iii) $\tan A = \sqrt{\sec^2 A - 1}$ (iv) $\operatorname{cosec} A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$ (v) $\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$