Answer
$\dfrac{\tan y}{\csc y}=\dfrac{1}{\cos y}-\dfrac{1}{\sec y}$
Work Step by Step
$\dfrac{\tan y}{\csc y}=\dfrac{1}{\cos y}-\dfrac{1}{\sec y}$
Substitute $\sec y$ with $\dfrac{1}{\cos y}$:
$\dfrac{\tan y}{\csc y}=\dfrac{1}{\cos y}-\dfrac{1}{\Big(\dfrac{1}{\cos y}\Big)}$
$\dfrac{\tan y}{\csc y}=\dfrac{1}{\cos y}-\cos y$
Evaluate the difference on the right side of the equation:
$\dfrac{\tan y}{\csc y}=\dfrac{1-\cos^{2}y}{\cos y}$
Since $1-\cos^{2}y=\sin^{2}y$, the equation becomes:
$\dfrac{\tan y}{\csc y}=\dfrac{\sin^{2}y}{\cos y}$
Rewrite the expression on the right side of the equation:
$\dfrac{\tan y}{\csc y}=\dfrac{\sin y}{\cos y}\dfrac{\sin y}{1}$
$\dfrac{\tan y}{\csc y}=\tan y\sin y$
Since $\sin y=\dfrac{1}{\csc y}$, the identity is proved:
$\dfrac{\tan y}{\csc y}=\dfrac{\tan y}{\csc y}$
