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, $$A=\cot\theta-\tan\theta$$
