Question

without tables, evaluate \\(\tan(\alpha + \beta)\\) and \\(\tan(\alpha - \beta)\\). show your work!

1. \\(\tan \alpha = \frac{4}{3}\\), \\(\cos \beta = \frac{12}{13}\\); \\(0 < \alpha < \frac{\pi}{2}\\), \\(0 < \beta < \frac{\pi}{2}\\)

Explanation:

Step1: Find \(\tan\beta\)

Given \(\cos\beta=\frac{12}{13}\) and \(0 < \beta<\frac{\pi}{2}\), we use the identity \(\sin^{2}\beta+\cos^{2}\beta = 1\) to find \(\sin\beta\).
\(\sin\beta=\sqrt{1-\cos^{2}\beta}=\sqrt{1 - (\frac{12}{13})^{2}}=\sqrt{\frac{169 - 144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}\) (since \(\beta\) is in the first quadrant, \(\sin\beta>0\)).
Then \(\tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}\).

Step2: Use the tangent addition formula for \(\tan(\alpha+\beta)\)

The formula for \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\). Here \(A=\alpha\), \(B = \beta\), \(\tan\alpha=\frac{4}{3}\), \(\tan\beta=\frac{5}{12}\).
Substitute the values: \(\tan(\alpha+\beta)=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times\frac{5}{12}}\)
First, simplify the numerator: \(\frac{4}{3}+\frac{5}{12}=\frac{16 + 5}{12}=\frac{21}{12}=\frac{7}{4}\)
Simplify the denominator: \(1-\frac{20}{36}=1-\frac{5}{9}=\frac{4}{9}\)
Then \(\tan(\alpha+\beta)=\frac{\frac{7}{4}}{\frac{4}{9}}=\frac{7}{4}\times\frac{9}{4}=\frac{63}{16}\)

Step3: Use the tangent subtraction formula for \(\tan(\alpha-\beta)\)

The formula for \(\tan(A - B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\). Here \(A=\alpha\), \(B=\beta\).
Substitute the values: \(\tan(\alpha - \beta)=\frac{\frac{4}{3}-\frac{5}{12}}{1+\frac{4}{3}\times\frac{5}{12}}\)
Simplify the numerator: \(\frac{4}{3}-\frac{5}{12}=\frac{16-5}{12}=\frac{11}{12}\)
Simplify the denominator: \(1+\frac{20}{36}=1+\frac{5}{9}=\frac{14}{9}\)
Then \(\tan(\alpha - \beta)=\frac{\frac{11}{12}}{\frac{14}{9}}=\frac{11}{12}\times\frac{9}{14}=\frac{99}{168}=\frac{33}{56}\)

Answer:

\(\tan(\alpha + \beta)=\frac{63}{16}\), \(\tan(\alpha-\beta)=\frac{33}{56}\)