$\mathrm{p}=\sin 2 \alpha, \mathrm{q}=\sin 2 \beta, \mathrm{r}=\cos 2 \alpha$, $\mathrm{s}=\cos 2 \beta \quad \mathrm{p}+\mathrm{q}=\mathrm{c}, \mathrm{r}+\mathrm{s}=\mathrm{d}$ হলে , $\cos (2 \alpha+2 \beta)=?$

Solve: দেওয়া আছে, $p=\sin 2 \alpha, q=\sin 2 \beta$,

$\begin{array}{rl} r & r=\cos 2 \alpha, s=\cos 2 \beta \\ \therefore & p+q=c \Rightarrow \sin 2 \alpha+\sin 2 \beta=c \cdot(i) \\ & r+s=d \Rightarrow \cos 2 \alpha+\cos 2 \beta=d \\ & \text { (ii) } \div \text { (i) } \Rightarrow \frac{\cos 2 \alpha+\cos 2 \beta}{\sin 2 \alpha+\sin 2 \beta}=\frac{d}{c} \\ \Rightarrow & \frac{2 \cos (\alpha+\beta) \cos (\alpha-\beta)}{2 \sin (\alpha+\beta) \cos (\alpha-\beta)}=\frac{d}{c} \\ \Rightarrow & \frac{\cos (\alpha+\beta)}{\sin (\alpha+\beta)}=\frac{d}{c} \Rightarrow \frac{\cos ^{2}(\alpha+\beta)}{\sin ^{2}(\alpha+\beta)}=\frac{d^{2}}{c^{2}} \\ \Rightarrow & \frac{\cos ^{2}(\alpha+\beta)+\sin ^{2}(\alpha+\beta)}{\cos ^{2}(\alpha+\beta)-\sin ^{2}(\alpha+\beta)}=\frac{d^{2}+c^{2}}{d^{2}-c^{2}} \end{array}$

[যোজন-বিয়োজন করে।]

$\begin{array}{l} \Rightarrow \frac{1}{\cos 2(\alpha+\beta)}=\frac{d^{2}+c^{2}}{d^{2}-c^{2}} \\ \therefore \cos (2 \alpha+2 \beta)=\frac{d^{2}-c^{2}}{d^{2}+c^{2}} \text { (Proved) } \end{array}$