Use the identity: $\sin^2 A + \cos^2 A = 1$ to prove that $\tan^2 A + 1 = \sec^2 A$. Hence, find the value of $\tan A$, when $\sec A = \frac{5}{3}$, where A is an acute angle.
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
Proof:
Starting with $\sin^2 A + \cos^2 A = 1$, divide both sides by $\cos^2 A$:
$$\frac{\sin^2 A}{\cos^2 A} + \frac{\cos^2 A}{\cos^2 A} = \frac{1}{\cos^2 A}$$
$$\tan^2 A + 1 = \sec^2 A \quad \textbf{(Proved)}$$
Finding tan A when sec A = 5/3:
$$\tan^2 A = \sec^2 A - 1 = \left(\frac{5}{3}\right)^2 - 1 = \frac{25}{9} - 1 = \frac{16}{9}$$
Since A is acute, $\tan A = \dfrac{4}{3}$.
Source: Chapter 8, Section 8.5 Summary
---
Explanation
- The proof requires dividing the Pythagorean identity by $\cos^2 A$ — this is the key step examiners look for.
- Always state "since A is acute, tan A is positive" to justify taking the positive square root; this earns the final mark.
- sec A = 5/3 > 1, which is valid (sec A ≥ 1 for acute angles).