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