Answer
$\dfrac{\sec\theta-\cos\theta}{\sin\theta}=\tan\theta$
Work Step by Step
$\dfrac{\sec\theta-\cos\theta}{\sin\theta}$
Substitute $\sec\theta$ with $\dfrac{1}{\cos\theta}$:
$\dfrac{\sec\theta-\cos\theta}{\sin\theta}=\dfrac{\dfrac{1}{\cos\theta}-\cos\theta}{\sin\theta}=...$
Evaluate the sum present in the numerator:
$...=\dfrac{\dfrac{1-\cos^{2}\theta}{\cos\theta}}{\sin\theta}=\dfrac{1-\cos^{2}\theta}{(\sin\theta)(\cos\theta)}=...$
We know that $\sin^{2}\theta+\cos^{2}\theta=1$, so if we solve for $\sin^{2}\theta$ here, we get the expression present in the numerator. So substituting $1-\cos^{2}\theta$ with $\sin^{2}\theta$, the expression becomes:
$...=\dfrac{\sin^{2}\theta}{(\sin\theta)(\cos\theta)}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta$
