$\frac{4}{3}$
We are given the equation \(\tan A + \sec A = 3\) and need to find the value of \(\tan A\).
We will use the fundamental trigonometric identity relating \(\sec A\) and \(\tan A\):
\(\sec^2 A - \tan^2 A = 1\)
This identity can be factored as a difference of squares:
\((\sec A - \tan A)(\sec A + \tan A) = 1\)
Substitute the given value: We know that \(\sec A + \tan A = 3\). Substitute this into the factored identity:
\((\sec A - \tan A)(3) = 1\)
Find \(\sec A - \tan A\): Divide both sides by 3:
\(\sec A - \tan A = \frac{1}{3}\)
Set up a system of equations: Now we have two equations:
Solve for \(\tan A\): Subtract Equation 2 from Equation 1 to eliminate \(\sec A\):
\((\sec A + \tan A) - (\sec A - \tan A) = 3 - \frac{1}{3}\)
\(\sec A + \tan A - \sec A + \tan A = \frac{9}{3} - \frac{1}{3}\)
\(2 \tan A = \frac{8}{3}\)
Calculate the final value: Divide by 2:
\(\tan A = \frac{8}{3} \times \frac{1}{2}\)
\(\tan A = \frac{4}{3}\)
The value of \(\tan A\) is \(\frac{4}{3}\).
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