## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Review Exercises - Page 248: 11

#### Answer

$$\tan\theta-\sec\theta\csc\theta=-\frac{\cos\theta}{\sin\theta}$$

#### Work Step by Step

$$A=\tan\theta-\sec\theta\csc\theta$$ - For $\tan\theta$: $$\tan\theta=\frac{\sin\theta}{\cos\theta}\hspace{1.5cm}\text{Quotient Identity}$$ - For $\sec\theta$: $$\sec\theta=\frac{1}{\cos\theta}\hspace{1.5cm}\text{Reciprocal Identity}$$ - For $\csc\theta$: $$\csc\theta=\frac{1}{\sin\theta}\hspace{1.5cm}\text{Reciprocal Identity}$$ Back to $A$: $$A=\frac{\sin\theta}{\cos\theta}-\frac{1}{\cos\theta}\times\frac{1}{\sin\theta}$$ $$A=\frac{\sin\theta}{\cos\theta}-\frac{1}{\sin\theta\cos\theta}$$ $$A=\frac{\sin^2\theta-1}{\sin\theta\cos\theta}$$ $$A=\frac{-(1-\sin^2\theta)}{\sin\theta\cos\theta}$$ - From Pythagorean Identities: $1-\sin^2\theta=\cos^2\theta$ $$A=\frac{-\cos^2\theta}{\sin\theta\cos\theta}$$ $$A=\frac{-\cos\theta}{\sin\theta}$$

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