Problem 43 - Precalculus Essentials

Question

Verify each identity. $$\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\sin x \cos y+\cos x \sin y}{\cos x \cos y-\sin x \sin y}$$

Step-by-Step Solution

Answer

After rewriting the right-side fraction using the tangential function, it becomes \(\frac{\tan(x) + \tan(y)}{1 - \tan(x) * \tan(y)}\), which matches the left side of the equation. Thus, the identity \(\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\sin x \cos y+\cos x \sin y}{\cos x \cos y-\sin x \sin y}\) has been verified.

Step 1: Rewrite in terms of tangent

Start by recalling that \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). So, rewrite the right-side fraction as \(\frac{\frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(y)}}{1 - \frac{\sin(x)}{\cos(x)} * \frac{\sin(y)}{\cos(y)}}\) = \(\frac{\tan(x) + \tan(y)}{1 - \tan(x) * \tan(y)}\).

Step 2: Use the trigonometric identity

Recall that \(\sin(x) \sin(y) + \cos(x) \cos(y)\) can be rewritten using the addition formula as \(\cos(x - y)\). Similarly, \(\cos(x) \cos(y) - \sin(x) \sin(y)\) can be rewritten as \(\cos(x + y)\). So, the right-side fraction becomes: \(\frac{\cos(x - y)}{\cos(x + y)}\).

Step 3: Check the identity

Check to see if you get the same value on both sides of the equation. If the left side equals the right side, then the identity has been verified.