Solution:

Let, $1+i=r(\cos \theta+i \sin \theta)=r \cos \theta+i \sin \theta$ Equating real \& imaginary parts on both sides,

$$ r \cos \theta=1 \quad \& \quad r \sin \theta=1\\ $$

Now,

$ (r \cos \theta)^{2}+(r \sin \theta)^{2}=(1)^{2}+(1)^{2}\\ $

Now,

$ (\operatorname{rcos} \theta)^{2}+(\operatorname{rsin} \theta)^{2}=(1)^{2}+(1)^{2} \\ $

$ \Rightarrow \mathrm{r}^{2} \cos ^{2} \theta+\mathrm{r}^{2} \sin ^{2} \theta=1+1 \\ $

$ \Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2 \\ $

$ \Rightarrow \quad \mathrm{r}^{2}(1)=2 \\ $

$ \Rightarrow r^{2}=2 \quad \Rightarrow r=\sqrt{2} \\ $

Also, $ \frac{r \sin \theta}{r \cos \theta}=\frac{1}{1} \\ $

$ \begin{aligned} &\Rightarrow \tan \theta=1 \\\\ &\Rightarrow \theta=\tan ^{-1}(1) \\\\ &\Rightarrow \theta=\frac{\pi}{4} \\\\ &\therefore 1+i=\sqrt{2}\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right]\\ \end{aligned} $

Similarly we can prove that, $$ 1-i=\sqrt{2}\left[\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right] \\ $$

LHS: $ (1+i)^{\mathrm{n}}+(1-i)^{\mathrm{n}}\\ $

$ \begin{aligned} &=\left[\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right]^{n}+\left[\sqrt{2}\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)\right]^{n}\right. \\\\ &=(\sqrt{2})^{n}\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right]^{n}+(\sqrt{2})^{n}\left[\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right]^{n} \\\\ &=(\sqrt{2})^{n}\left[\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}+\cos \frac{n \pi}{4}-i \sin \frac{n \pi}{4}\right] \\\\ &=2^{\frac{n}{2}} 2 \cos \frac{n \pi}{4} \\\\ &=2^{\frac{n}{2}+1} \cos \frac{n \pi}{4} \\\\ &=2^{\frac{n+2}{2}} \cos \frac{n \pi}{4}=\text { RHS }\\ \end{aligned} $