Answer
$a)$ $\dfrac{\tan y}{\csc y}=\sec y-\cos y$
$b)$
Work Step by Step
$\dfrac{\tan y}{\csc y}=\sec y-\cos y$
$a)$ Verify algebraically
On the left side, substitute $\tan y$ by $\dfrac{\sin y}{\cos y}$ and $\csc y$ by $\dfrac{1}{\sin x}$:
$\dfrac{\Big(\dfrac{\sin y}{\cos y}\Big)}{\Big(\dfrac{1}{\sin y}\Big)}=\sec y-\cos y$
Evaluate the division on the left side:
$\dfrac{(\sin y)(\sin y)}{\cos y}=\sec y-\cos y$
$\dfrac{\sin^{2}y}{\cos y}=\sec y-\cos y$
Substitute $\sin^{2}y$ by $1-\cos^{2}y$:
$\dfrac{1-\cos^{2}y}{\cos y}=\sec y-\cos y$
Rewrite $\dfrac{1-\cos^{2}y}{\cos y}$ as $\dfrac{1}{\cos y}-\dfrac{\cos^{2}y}{\cos y}$ and simplify:
$\dfrac{1}{\cos y}-\dfrac{\cos^{2}y}{\cos y}=\sec y-\cos y$
$\dfrac{1}{\cos y}-\cos y=\sec y-\cos y$
Since $\dfrac{1}{\cos y}=\sec y$, the identity is proved.
$b)$ Confirm graphically
Graph both sides of the equation. Since this identity is true, both graphs must be the same
