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Chapter 6: Problem 56

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.

Short Answer

Expert verified
The identity for \( \tan (\alpha-\beta) \) is \( \tan (\alpha-\beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\).

Step by step solution

01

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: \(\tan (\alpha-\beta)\) = \(\frac{\tan(\alpha) + \tan(-\beta)}{1 - \tan(\alpha)\tan(-\beta)}\).
02

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: \(\tan (\alpha-\beta)\) = \(\frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\).
03

Final Step

The identity for \( \tan (\alpha-\beta) \) is therefore \( \tan (\alpha-\beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\).

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