write each expression in terms of sine and cosine, and then simplify so that no quotients appear in the…

Explanation:

Step1: Substitute trigonometric identities

We know that (\csc\theta=\frac{1}{\sin\theta}) and (\sec\theta = \frac{1}{\cos\theta}). So the expression ((\csc\theta+\sec\theta)(\sin\theta - \cos\theta)) becomes (\left(\frac{1}{\sin\theta}+\frac{1}{\cos\theta}\right)(\sin\theta-\cos\theta)).

Step2: Simplify the sum of fractions

First, simplify (\frac{1}{\sin\theta}+\frac{1}{\cos\theta}=\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}). Then the expression is (\frac{\sin\theta + \cos\theta}{\sin\theta\cos\theta}\times(\sin\theta-\cos\theta)).

Step3: Apply the difference - of - squares formula ((a + b)(a - b)=a^{2}-b^{2})

Here (a=\sin\theta) and (b = \cos\theta), so (\frac{\sin^{2}\theta-\cos^{2}\theta}{\sin\theta\cos\theta}).

Step4: Use double - angle formulas

We know that (\sin^{2}\theta-\cos^{2}\theta=-\cos(2\theta)) and (\sin\theta\cos\theta=\frac{1}{2}\sin(2\theta)). But we can also rewrite it as follows: [ \begin{align*} \frac{\sin^{2}\theta-\cos^{2}\theta}{\sin\theta\cos\theta}&=\frac{\sin^{2}\theta}{\sin\theta\cos\theta}-\frac{\cos^{2}\theta}{\sin\theta\cos\theta}\ &=\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta}\ &=\tan\theta-\cot\theta \end{align*} ] Another way: [ \begin{align*} \left(\frac{1}{\sin\theta}+\frac{1}{\cos\theta}\right)(\sin\theta - \cos\theta)&=\frac{1}{\sin\theta}\times\sin\theta-\frac{1}{\sin\theta}\times\cos\theta+\frac{1}{\cos\theta}\times\sin\theta-\frac{1}{\cos\theta}\times\cos\theta\ &=1-\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}-1\ &=\frac{\sin^{2}\theta-\cos^{2}\theta}{\sin\theta\cos\theta}\ &=\tan\theta-\cot\theta \end{align*} ]

Answer:

(\tan\theta-\cot\theta)