Answer
$\dfrac{g(x+h)-g(x)}{h}=(\dfrac{\sin h}{h})\cos x-\sin x\dfrac{(1-\cos h)}{h}$
Work Step by Step
Use the identity$\sin (a+b)=\sin a \cos b+\cos a\sin b$
Let us consider $g(x)=\sin x$
and $g(x+h)=\sin (x+h)=\sin x \cos h+\cos x \sin h$
Now, $\dfrac{g(x+h)-g(x)}{h}=\dfrac{\sin x \cos h+\cos x \sin h-\sin x}{h}$
or, $=\dfrac{-\sin x(1-\cos h)+\sin h \cos x}{h}$
or, $\dfrac{g(x+h)-g(x)}{h}=(\dfrac{\sin h}{h})\cos x-\sin x\dfrac{(1-\cos h)}{h}$
Hence, the result has been proved.
