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: $1-\sin^2\theta=\cos^2\theta$
$$A=\frac{-\cos^2\theta}{\sin\theta\cos\theta}$$
$$A=\frac{-\cos\theta}{\sin\theta}$$
