If sin α sin β − cos α cos β + 1 = 0,

Given:

$\sin \alpha s$ in $\beta-\cos \alpha \cos \beta+1=0$

$\Rightarrow-(\cos \alpha \cos \beta-\sin \alpha \sin \beta)+1=0$

$\Rightarrow-\cos (\alpha+\beta)+1=0$

$\Rightarrow \cos (\alpha+\beta)=1$

Therefore, $\sin (\alpha+\beta)=0 \quad \ldots(1)\left(\right.$ Since $\left.\sin \theta=\sqrt{1-\cos ^{2} \theta}\right)$

Hence ,

$1+\cot \alpha \tan \beta=1+\frac{\cos \alpha \sin \beta}{\sin \alpha \cos \beta}$

$=\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \cos \beta}$

$=\frac{\sin (\alpha+\beta)}{\sin \alpha \cos \beta}$

$=0 \quad \ldots\{$ From eq (1) $\}$

Hence proved.