Problem 47 - Precalculus Essentials

Question

Verify each identity. $$\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{\tan \alpha+\tan \beta}{\tan \alpha-\tan \beta}$$

Step-by-Step Solution

Answer

The identity \( \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{\tan \alpha+\tan \beta}{\tan \alpha-\tan \beta} \) is verified.

Step 1: Apply Addition and Subtraction Formulas for Sine

First recall the addition and subtraction formulas for sine:\( \sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta \)\( \sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta \)Substitute these formulas into the original expression, such that the left side of the equation becomes:\( \frac{\sin\alpha \cos\beta + \cos\alpha \sin\beta}{\sin\alpha \cos\beta - \cos\alpha \sin\beta} \)

Step 2: Rewrite Sine in Terms of Tangent

Now, express \( \sin\alpha \) and \( \sin\beta \) as \( \frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}} \) and \( \frac{\tan\beta}{\sqrt{1+\tan^2\beta}} \) respectively, and \( \cos\alpha \) and \( \cos\beta \) as \( \frac{1}{\sqrt{1+\tan^2\alpha}} \) and \( \frac{1}{\sqrt{1+\tan^2\beta}} \), respectively. Substituting these expressions into the equation from Step 1, we have:\( \frac{\frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}}\frac{1}{\sqrt{1+\tan^2\beta}}+\frac{1}{\sqrt{1+\tan^2\alpha}}\frac{\tan\beta}{\sqrt{1+\tan^2\beta}}}{\frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}}\frac{1}{\sqrt{1+\tan^2\beta}}-\frac{1}{\sqrt{1+\tan^2\alpha}}\frac{\tan\beta}{\sqrt{1+\tan^2\beta}}} \)

Step 3: Simplify the Expression

Further simplifying we get \( \frac{\tan\alpha+\tan\beta}{\tan\alpha-\tan\beta} \)The result is identical with the right hand side of the original identity, which confirms it.