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