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Chapter 5: Problem 9

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

Short Answer

Expert verified
The given identity \(\frac{\cos (\alpha-\beta)}{\cos \alpha \sin \beta}=\tan \alpha + \cot \beta\) is verified as true as the softmax function can be broken down into two parts \(\frac{\cos \alpha \cos \beta}{\cos \alpha \sin \beta} + \frac{\sin \alpha \sin \beta}{\cos \alpha \sin \beta}\) and simplified to \(\tan \alpha + \cot \beta\).

Step by step solution

01

Write down the given expression

The given expression is \(\frac{\cos (\alpha-\beta)}{\cos \alpha \sin \beta}\). The target is to show that this expression equals to \(\tan \alpha + \cot \beta\).
02

Use Addition and Subtraction Formulas

The cosine of a difference identity \(\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\). Substitute this back into the original equation.
03

Simplify the new expression

The expression after substitution is \(\frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\cos \alpha \sin \beta}\). This can be rewritten as \(\frac{\cos \alpha \cos \beta}{\cos \alpha \sin \beta} + \frac{\sin \alpha \sin \beta}{\cos \alpha \sin \beta}\).
04

Simplify further

The expression then boils down to \(\frac{\cos \beta}{\sin \beta} + \frac{\sin \alpha}{\cos \alpha}\). This can be rewritten in terms of the reciprocal trigonometric functions, cotangent and tangent, as \(\cot \beta + \tan \alpha\).
05

Conclusion

Hence, the original expression \(\frac{\cos (\alpha-\beta)}{\cos \alpha \sin \beta}\) is equivalent to \(\tan \alpha + \cot \beta\), thereby verifying the identity

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