Answer
See the full explanation below.
Work Step by Step
Let us consider the left side of the given expression:
$\frac{\cos \,\left( x+h \right)-\cos \,x}{h}$
By using the identity of trigonometry,
$\cos \,\left( \alpha +\beta \right)=\cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta $ ,
Now, the above given expression can be further simplified as:
$\begin{align}
& \frac{\cos \,\left( x+h \right)-\cos \,x}{h}=\frac{\cos \,x\,\cos \,h-\sin \,x\,\sin \,h-\cos \,x}{h} \\
& =\frac{\cos \,x\,\cos \,h-\cos \,x-\sin \,x\,\sin \,h}{h} \\
& =\frac{\cos \,x\left( \,\cos \,h-1 \right)-\sin \,x\,\sin \,h}{h} \\
& =\cos \,x\frac{\left( \,\cos \,h-1 \right)}{h}-\sin \,x\,\frac{\sin \,h}{h}
\end{align}$
Hence, the left side of the expression is equal to the right side, which is $\frac{\cos \,\left( x+h \right)-\cos \,x}{h}=\cos \,x\frac{\cos \,h-1}{h}-\sin \,x\frac{\sin \,h}{h}$.
