Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Step-by-Step Solution
Key Concept: Use the definition of cotangent $\cot A = \dfrac{\cos A}{\sin A}$ (or $\cot A = \dfrac{1}{\tan A}$) together with the fundamental identity $\sin^2 A + \cos^2 A = 1$ to rewrite the required ratios solely in terms of $\cot A$.
1. Write the definition of cotangent:
$$\cot A = \frac{\cos A}{\sin A} \quad\Rightarrow\quad \cos A = \cot A \cdot \sin A.$$
2. Use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$.
Substitute $\cos A = \cot A \; \sin A$:
$$\sin^2 A + (\cot A \; \sin A)^2 = 1$$
$$\sin^2 A \bigl(1 + \cot^2 A\bigr) = 1.$$
3. Solve for $\sin A$ (for an acute angle $A$, $\sin A>0$):
$$\sin A = \frac{1}{\sqrt{1+\cot^2 A}}.$$
4. Find $\cos A$ using $\cos A = \cot A \; \sin A$:
$$\cos A = \cot A \cdot \frac{1}{\sqrt{1+\cot^2 A}} = \frac{\cot A}{\sqrt{1+\cot^2 A}}.$$
5. Express $\tan A$ using $\tan A = \dfrac{\sin A}{\cos A}$ or directly from $\cot A = \dfrac{1}{\tan A}$:
$$\tan A = \frac{1}{\cot A}.$$
6. Express $\sec A$ as the reciprocal of $\cos A$:
$$\sec A = \frac{1}{\cos A} = \frac{\sqrt{1+\cot^2 A}}{\cot A}.$$
7. Summarise the results:
$$\boxed{\sin A = \frac{1}{\sqrt{1+\cot^2 A}}},\qquad \boxed{\sec A = \frac{\sqrt{1+\cot^2 A}}{\cot A}},\qquad \boxed{\tan A = \frac{1}{\cot A}}.$$
Correct Answer: sin A = 1/√(1+cot²A), sec A = √(1+cot²A)/cot A, tan A = 1/cot A