Question

Use the identity: $\sin^2A + \cos^2A = 1$ to prove that $\tan^2A + 1 = \sec^2A$. Hence, find the value of $\tan A$, when $\sec A = \dfrac{5}{3}$ where A is an acute angle.

Hint:

Always square carefully when working with reciprocal trigonometric identities.

Answer 1

Problem:
Given that \( \sec A = \dfrac{5}{3} \), calculate \( \tan A \)

Step 1: Apply Pythagorean identity
Recall the trigonometric identity:\[\tan^2 A + 1 = \sec^2 A\]
Step 2: Substitute \( \sec A \)
\[\sec A = \frac{5}{3} \Rightarrow \sec^2 A = \left( \frac{5}{3} \right)^2 = \frac{25}{9}\]
Substitute into the identity:\[\tan^2 A + 1 = \frac{25}{9}\Rightarrow \tan^2 A = \frac{25}{9} - 1= \frac{25 - 9}{9} = \frac{16}{9}\]
Step 3: Solve for \( \tan A \)
\[\tan A = \sqrt{\frac{16}{9}} = \frac{4}{3}\]
(Note: Positive root is used because no quadrant is specified and values are positive.)

Final Answer:
\[\boxed{\tan A = \dfrac{4}{3}}\]