Problem 48 - Precalculus Essentials

Question

Verify each identity. $$\frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{1-\tan \alpha \tan \beta}{1+\tan \alpha \tan \beta}$$

Step-by-Step Solution

Answer

The given trigonometric identity has been verified correctly.

Step 1: Formula for cosine of a sum and difference

First, recall the trigonometric formula for the cosine of the sum and difference of two angles: \[ \cos (A + B) = \cos A \cos B - \sin A \sin B \] \[ \cos (A - B) = \cos A \cos B + \sin A \sin B \]

Step 2: Apply the formulas for the identity

Plugging these formulas into our identity gives: \[ \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta }\ = \frac{1 - \tan \alpha \tan \beta}{1 + \tan \alpha \tan \beta} \]

Step 3: Convert to tangent

Express sin and cos as the division of sin over cos, which gives the tangent (\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha}\]), we obtain: \[ \frac{1 - \frac{\sin \alpha}{\cos \alpha} \frac{\sin \beta}{\cos \beta}}{1 + \frac{\sin \alpha}{\cos \alpha} \frac{\sin \beta}{\cos \beta}}\]\[ \Rightarrow \frac{1 - \tan \alpha \tan \beta}{1 + \tan \alpha \tan \beta} \]

Step 4: Verify identity

With this, we have transformed left-hand side to look exactly like right-hand side, so the identity is verified.