Answer
$$(1-\cos\theta)(1+\sec\theta)=\sin\theta\tan\theta$$
Work Step by Step
$$A=(1-\cos\theta)(1+\sec\theta)$$
- Reciprocal Identity: $$\sec\theta=\frac{1}{\cos\theta}$$
Replace into $A$:
$$A=(1-\cos\theta)(1+\frac{1}{\cos\theta})$$
$$A=(1-\cos\theta)(\frac{\cos\theta+1}{\cos\theta})$$
$$A=\frac{(1-\cos\theta)(1+\cos\theta)}{\cos\theta}$$
As $(a-b)(a+b)=a^2-b^2$:
$$A=\frac{1-\cos^2\theta}{\cos\theta}$$
- Pythagorean Identity:
$$\sin^2\theta=1-\cos^2\theta$$
Replace into $A$:
$$A=\frac{\sin^2\theta}{\cos\theta}$$
As the exercise requests no quotients appear in the final expression, we continue.
$$A=\sin\theta\times\frac{\sin\theta}{\cos\theta}$$
$$A=\sin\theta\tan\theta\hspace{1cm}\text{(Quotient Identity)}$$
