Q2:

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

Given: An angle $A$ in a right-angled triangle, where the trigonometric ratio is expressed in terms of $\sec A$.

To Find: Express $\sin A$, $\cos A$, $\tan A$, $\csc A$, and $\cot A$ in terms of $\sec A$.

Step 1: Expressing $\cos A$ in terms of $\sec A$

By the definition of reciprocal trigonometric ratios, we know that $\cos A$ is the reciprocal of $\sec A$.

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

Step 2: Expressing $\sin A$ in terms of $\sec A$

Using the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$.

$\sin^2 A = 1 - \cos^2 A$ [Subtracting $\cos^2 A$ from both sides]

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

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

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

Step 3: Expressing $\tan A$ in terms of $\sec A$

Using the identity: $1 + \tan^2 A = \sec^2 A$.

$\tan^2 A = \sec^2 A - 1$ [Subtracting 1 from both sides]

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

Step 4: Expressing $\csc A$ in terms of $\sec A$

By definition, $\csc A = \frac{1}{\sin A}$.

$\csc A = \frac{1}{\frac{\sqrt{\sec^2 A - 1}}{\sec A}}$ [Substituting the expression for $\sin A$ derived in Step 2]

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

Step 5: Expressing $\cot A$ in terms of $\sec A$

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

$\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$ [Substituting the expression for $\tan A$ derived in Step 3]

Final Answer:

The trigonometric ratios in terms of $\sec A$ are:

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

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

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

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

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