Problem 56

Question

Derive the identity for \(\tan (\alpha-\beta)\) using $$ \tan (\alpha-\beta)=\tan [\alpha+(-\beta)] $$ After applying the formula for the tangent of the sum of two angles, use the fact that the tangent is an odd function.

Step-by-Step Solution

Verified
Answer
The identity for \( \tan (\alpha-\beta) \) is \( \tan (\alpha-\beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\).
1Step 1: Use the formula for the tangent of the sum of two angles
Plug \(\alpha\) and \(-\beta\) into the formula for the sum of two angles, giving us: \(\tan (\alpha-\beta)\) = \(\frac{\tan(\alpha) + \tan(-\beta)}{1 - \tan(\alpha)\tan(-\beta)}\).
2Step 2: Recognize \(\tan\) as an odd function
Replace \(\tan(-\beta)\) with \(-\tan(\beta)\), using the property that the tangent function is an odd function. This gives: \(\tan (\alpha-\beta)\) = \(\frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\).
3Step 3: Final Step
The identity for \( \tan (\alpha-\beta) \) is therefore \( \tan (\alpha-\beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\).