L.H.S. $= \sin(n+1)x \sin(n+2)x + \cos(n+1)x \cos(n+2)x$

We can write $(n+2)x$ as $[(n+1)x + x]$. Substituting this into the expression, we get:

L.H.S. $= \sin(n+1)x \sin[(n+1)x + x] + \cos(n+1)x \cos[(n+1)x + x]$

Using the expansion formulas from the context, $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$ (as seen in the derivation for $\cos(2x+x)$), let $A = (n+1)x$ and $B = x$.

Expanding the terms:

$$\sin[(n+1)x + x] = \sin(n+1)x \cos x + \cos(n+1)x \sin x$$ $$\cos[(n+1)x + x] = \cos(n+1)x \cos x - \sin(n+1)x \sin x$$

Now, substitute these back into the LHS expression:

L.H.S. $= \sin(n+1)x [\sin(n+1)x \cos x + \cos(n+1)x \sin x] + \cos(n+1)x [\cos(n+1)x \cos x - \sin(n+1)x \sin x]$

Multiplying the terms:

$$= \sin^2(n+1)x \cos x + \sin(n+1)x \cos(n+1)x \sin x + \cos^2(n+1)x \cos x - \cos(n+1)x \sin(n+1)x \sin x$$

Grouping like terms:

$$= [\sin^2(n+1)x + \cos^2(n+1)x] \cos x + [\sin(n+1)x \cos(n+1)x - \cos(n+1)x \sin(n+1)x] \sin x$$

Since $\sin^2 \theta + \cos^2 \theta = 1$ and the coefficient of $\sin x$ cancels out to zero, we get:

$$= [1] \cos x + [0] \sin x$$ $$= \cos x$$ $$= \textbf{R.H.S.}$$

Hence proved.