Prove that

Solution:

To Prove: $\frac{\cos ^{3} x-\sin ^{3} x}{\cos x-\sin x}=\frac{1}{2}(2+\sin 2 x)$

Taking LHS,

$=\frac{\cos ^{3} x-\sin ^{3} x}{\cos x-\sin x} \ldots$ (i)

We know that,

$a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$

So, $\cos ^{3} x-\sin ^{3} x=(\cos x-\sin x)\left(\cos ^{2} x+\cos x \sin x+\sin ^{2} x\right)$

So, eq. (i) becomes

$=\frac{(\cos x-\sin x)\left(\cos ^{2} x+\cos x \sin x+\sin ^{2} x\right)}{\cos x-\sin x}$

$=\cos ^{2} x+\cos x \sin x+\sin ^{2} x$

$=\left(\cos ^{2} x+\sin ^{2} x\right)+\cos x \sin x$

$=(1)+\cos x \sin x\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$=1+\cos x \sin x$

Multiply and Divide by 2, we get

$=\frac{1}{2}[2 \times(1+\cos x \sin x)]$

$=\frac{1}{2}[2+2 \sin x \cos x]$

$=\frac{1}{2}[2+\sin 2 x]$ [∵ sin 2x = 2 sinx cosx]

= RHS

∴ LHS = RHS

Hence Proved