## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 201: 69

#### Answer

$$(\sec\theta+\csc\theta)(\cos\theta-\sin\theta)=\cot\theta-\tan\theta$$

#### Work Step by Step

$$A=(\sec\theta+\csc\theta)(\cos\theta-\sin\theta)$$ - Reciprocal Identities: $$\sec\theta=\frac{1}{\cos\theta}$$ $$\csc\theta=\frac{1}{\sin\theta}$$ Replace into $A$: $$A=(\frac{1}{\cos\theta}+\frac{1}{\sin\theta})(\cos\theta-\sin\theta)$$ $$A=\Big(\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}\Big)(\cos\theta-\sin\theta)$$ $$A=\frac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}{\sin\theta\cos\theta}$$ $$A=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}$$ (for $(A-B)(A+B)=A^2-B^2$) Now we can separate the numerator to eliminate the denominator: $$A=\frac{\cos^2\theta}{\sin\theta\cos\theta}-\frac{\sin^2\theta}{\sin\theta\cos\theta}$$ $$A=\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}$$ - Quotient Identities: $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$ $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ Therefore, $$A=\cot\theta-\tan\theta$$

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