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

We know: \[ \sin^2A + \cos^2A = 1 \Rightarrow \frac{\sin^2A}{\cos^2A} + 1 = \frac{1}{\cos^2A} \Rightarrow \tan^2A + 1 = \sec^2A \] Now, given $\sec A = \dfrac{5}{3}$, so: \[ \sec^2A = \left(\dfrac{5}{3}\right)^2 = \dfrac{25}{9}\\ \Rightarrow \tan^2A = \dfrac{25}{9} - 1 = \dfrac{25 - 9}{9} = \dfrac{16}{9}\\ \Rightarrow \tan A = \dfrac{4}{3} \] Answer: $\tan A = \dfrac{4}{3}$