Express the ratios cos A, tan A and sec A in terms of sin A.
Step-by-Step Solution
Key Concept: Use the fundamental Pythagorean identity $\sin^2 A + \cos^2 A = 1$ to write $\cos A$ in terms of $\sin A$, and then use the definitions $\tan A = \dfrac{\sin A}{\cos A}$ and $\sec A = \dfrac{1}{\cos A}$.
1. Pythagorean identity\
$$\sin^2 A + \cos^2 A = 1$$\
Rearranging,\
$$\cos^2 A = 1 - \sin^2 A.$$\
2. Express $\cos A$\
Since the example deals with acute angles ($0^\circ < A < 90^\circ$), $\cos A$ is positive. Hence,\
$$\boxed{\cos A = \sqrt{1 - \sin^2 A}}.$$\
3. Express $\tan A$\
By definition, $\tan A = \dfrac{\sin A}{\cos A}$. Substituting the expression for $\cos A$ obtained above,\
$$\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}.$$\
Therefore,\
$$\boxed{\tan A = \dfrac{\sin A}{\sqrt{1 - \sin^2 A}}}.$$\
4. Express $\sec A$\
By definition, $\sec A = \dfrac{1}{\cos A}$. Using the expression for $\cos A$,\
$$\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}.$$\
Hence,\
$$\boxed{\sec A = \dfrac{1}{\sqrt{1 - \sin^2 A}}}.$$\
Thus, all three required ratios are expressed solely in terms of $\sin A$.
Correct Answer: cos A = \sqrt{1 - \sin^2 A}, \; tan A = \dfrac{\sin A}{\sqrt{1 - \sin^2 A}}, \; sec A = \dfrac{1}{\sqrt{1 - \sin^2 A}}