#### Answer

$$\frac{1}{\sin\theta}$$

#### Work Step by Step

To simplify
$$\frac{\tan\theta}{\sec\theta-\cos\theta},$$
substitute the Reciprocal Identity
$$\sec\theta=\frac{1}{\cos\theta}$$
and the Tangent Identity
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
to obtain
$$\frac{\tan\theta}{\sec\theta-\cos\theta}=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}-\cos\theta}.$$
Multiply both numerator and denominator by $\cos\theta$
$$=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}-\cos\theta}\times \frac{\cos\theta}{\cos\theta}=\frac{\sin\theta}{1-\cos^{2}\theta}$$
Finally, use the Pythagorean Identity:
$$\sin^{2}\theta+\cos^{2}\theta=1$$
rearranged as
$$1-\cos^{2}\theta=\sin^{2}\theta$$
to substitute and obtain
$$\frac{\tan\theta}{\sec\theta-\cos\theta}=\frac{\sin\theta}{\sin^{2}\theta}=\frac{1}{\sin\theta}$$