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 \( \sec A = \dfrac{5}{3} \), find the value of \( \tan A \)

Step 1: Use the Pythagorean identity
We know from trigonometric identities that: \[ \tan^2 A + 1 = \sec^2 A \]
Step 2: Substitute the given value of \( \sec A \)
\[ \sec A = \frac{5}{3} \Rightarrow \sec^2 A = \left( \frac{5}{3} \right)^2 = \frac{25}{9} \]
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: Take square root of both sides
\[ \tan A = \sqrt{\frac{16}{9}} = \frac{4}{3} \]
(Note: Since no quadrant is specified and all values are positive, we consider the positive root.)

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