Write all the other trigonometric ratios of A in terms of sec A.
Step-by-Step Solution
Key Concept: Use the fundamental identity $\sin^2 A + \cos^2 A = 1$ and the definition $\sec A = \dfrac{1}{\cos A}$ to express $\cos A$, $\sin A$, $\tan A$, $\csc A$, and $\cot A$ solely in terms of $\sec A$.
1. Express $\cos A$ in terms of $\sec A$
$$\sec A = \frac{1}{\cos A} \;\Rightarrow\; \cos A = \frac{1}{\sec A}.$$
2. Find $\sin A$ using $\sin^2 A + \cos^2 A = 1$
\[
\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{1}{\sec A}\right)^2 = 1 - \frac{1}{\sec^2 A}.
\]
\[
\sin^2 A = \frac{\sec^2 A - 1}{\sec^2 A} \;\Rightarrow\; \sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}.
\]
(We take the positive root for acute angles; the sign can be adjusted according to the quadrant.)
3. Obtain $\tan A$
\[
\tan A = \frac{\sin A}{\cos A} = \frac{\dfrac{\sqrt{\sec^2 A - 1}}{\sec A}}{\dfrac{1}{\sec A}} = \sqrt{\sec^2 A - 1}.
\]
4. Find $\csc A$
\[
\csc A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}.
\]
5. Find $\cot A$
\[
\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}.
\]
Summary of all ratios in terms of $\sec A$
\[
\begin{aligned}
\cos A &= \frac{1}{\sec A},\\[4pt]
\sin A &= \frac{\sqrt{\sec^2 A - 1}}{\sec A},\\[4pt]
\tan A &= \sqrt{\sec^2 A - 1},\\[4pt]
\csc A &= \frac{\sec A}{\sqrt{\sec^2 A - 1}},\\[4pt]
\cot A &= \frac{1}{\sqrt{\sec^2 A - 1}}.
\end{aligned}
\]
Correct Answer: \cos A = \frac{1}{\sec A}, \; \sin A = \frac{\sqrt{\sec^2 A - 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}}.