If cos (α + β) sin (γ + δ) = cos (α − β) sin (γ − δ),

$\cos (\alpha+\beta) \sin (\gamma+\delta)=\cos (\alpha-\beta) \sin (\gamma-\delta)$

$\Rightarrow[\cos \alpha \cos \beta-\sin \alpha \sin \beta][\sin \gamma \cos \delta+\cos \gamma \sin \delta]=$ $[\cos \alpha \cos \beta+\sin \alpha \sin \beta][\sin \gamma \cos \delta-\cos \gamma \sin \delta]$

Dividing both sides by $\sin \alpha \sin \beta \sin \gamma \sin \delta:$

$\frac{[\cos \alpha \cos \beta-\sin \alpha \sin \beta][\sin \gamma \cos \delta+\cos \gamma \sin \delta]}{\sin \alpha \sin \beta \sin \gamma \sin \delta}=\frac{[\cos \alpha \cos \beta+\sin \alpha \sin \beta][\sin \gamma \cos \delta-\cos \gamma \sin \delta]}{\sin \alpha \sin \beta \sin \gamma \sin \delta}$

$\Rightarrow \frac{[\cos \alpha \cos \beta-\sin \alpha \sin \beta]}{\sin \alpha \sin \beta} \times \frac{[\sin \gamma \cos \delta+\cos \gamma \sin \delta]}{\sin \gamma \sin \delta}=\frac{[\cos \alpha \cos \beta+\sin \alpha \sin \beta]}{\sin \alpha \sin \beta} \times \frac{[\sin \gamma \cos \delta-\cos \gamma \sin \delta]}{\sin \gamma \sin \delta}$

$\Rightarrow[\cot \alpha \cot \beta-1][\cot \delta+\cot \gamma]=[\cot \alpha \cot \beta+1][\cot \delta-\cot \gamma]$

$\Rightarrow \cot \alpha \cot \beta \cot \delta+\cot \alpha \cot \beta \cot \gamma-\cot \delta-\cot \gamma=\cot \alpha \cot \beta \cot \delta-\cot \alpha \cot \beta \cot \gamma+\cot \delta-\cot \gamma$

$\Rightarrow-\cot \delta-\cot \delta=-\cot \alpha \cot \beta \cot \gamma-\cot \alpha \cot \beta \cot \gamma$

$\Rightarrow-2 \cot \delta=-2 \cot \alpha \cot \beta \cot \gamma$

$\Rightarrow \cot \alpha \cot \beta \cot \gamma=\cot \delta$

Hence proved.