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