$\begin{aligned} & \text { Given, } \alpha+\beta=\gamma \\ & \text { Then, } \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma \\ & =\frac{1}{2}\left[2 \cos ^2 \alpha+2 \cos ^2 \beta+2 \cos ^2 \gamma\right] \\ & \qquad=\frac{1}{2}\left[1+\cos 2 \alpha+1+\cos 2 \beta+2 \cos ^2 \gamma\right] \\ & \qquad\left[\because 2 \cos ^2 A=1+\cos 2 A\right] \\ & =\frac{1}{2}\left[2+\cos 2 \alpha+\cos 2 \beta+2 \cos ^2 \gamma\right] \\ & =\frac{1}{2}\left[2+2 \cos \left(\frac{2 \alpha+2 \beta}{2}\right) \cos \left(\frac{2 \alpha-2 \beta}{2}\right)+2 \cos ^2 \gamma\right] \\ & =\frac{1}{2}\left[2+2 \cos \left(\frac{2(\gamma)}{2}\right) \cos (\alpha-\beta)+2 \cos ^2 \gamma\right]\end{aligned}$
$\begin{aligned} & =\frac{1}{2}\left[2+2 \cos \gamma \cos (\alpha-\beta)+2 \cos ^2 \gamma\right] \\ & =1+\cos \gamma \cos (\alpha-\beta)+\cos ^2 \gamma \\ & =1+\cos \gamma\{\cos (\alpha-\beta)+\cos \gamma\} \\ & =1+\cos \gamma\left\{2 \cos \left(\frac{\alpha-\beta+\gamma}{2}\right) \cdot \cos \left(\frac{\alpha-\beta-\gamma}{2}\right)\right\} \\ & =1+2 \cos \gamma \cos \left(\frac{\alpha-\beta+\alpha+\beta}{2}\right) \cdot \cos \left(\frac{\alpha-\beta-\alpha-\beta}{2}\right) \\ & =1+2 \cos \gamma \cos \left(\frac{2 \alpha}{2}\right) \cdot \cos \left(-\frac{2 \beta}{2}\right) \\ & =1+2 \cos \gamma \cos \alpha \cdot \cos (-\beta) \\ & =1+2 \cos \alpha \cos \beta \cos \gamma \\ & \quad[\because \cos (-\theta)=\cos \theta]\end{aligned}$