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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.
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$\sin^2 A + \cos^2 A = 1$. Dividing both sides by $\cos^2 A$, we get $\frac{\sin^2 A}{\cos^2 A} + \frac{\cos^2 A}{\cos^2 A} = \frac{1}{\cos^2 A}$ ($\frac{1}{2}$ mark). $\tan^2 A + 1 = \sec^2 A$ ($\frac{1}{2}$ mark). $\tan^2 A + 1 = (\frac{5}{3})^2$ ($\frac{1}{2}$ mark). $\tan A = \frac{4}{3}$ ($\frac{1}{2}$ mark).