Question

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

Answer 1

$$\begin{gathered} \cos \left(\alpha +\beta \right)\sin \left(\gamma +\delta \right)=\cos \left(\alpha -\beta \right)\sin \left(\gamma -\delta \right) \\ \Rightarrow \left[\cos \alpha \cos \beta -\sin \alpha \sin \beta \right]\left[\sin \gamma \cos \delta +\cos \gamma \sin \delta \right]=\left[\cos \alpha \cos \beta +\sin \alpha \sin \beta \right]\left[\sin \gamma \cos \delta -\cos \gamma \sin \delta \right] \end{gathered}$$

$$\begin{gathered} \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sin \alpha \sin \beta \sin \gamma \sin \delta : \\ \frac{\left[\cos \alpha \cos \beta -\sin \alpha \sin \beta \right]\left[\sin \gamma \cos \delta +\cos \gamma \sin \delta \right]}{\sin \alpha \sin \beta \sin \gamma \sin \delta }=\frac{\left[\cos \alpha \cos \beta +\sin \alpha \sin \beta \right]\left[\sin \gamma \cos \delta -\cos \gamma \sin \delta \right]}{\sin \alpha \sin \beta \sin \gamma \sin \delta } \\ \Rightarrow \frac{\left[\cos \alpha \cos \beta -\sin \alpha \sin \beta \right]}{\sin \alpha \sin \beta }\times \frac{\left[\sin \gamma \cos \delta +\cos \gamma \sin \delta \right]}{\sin \gamma \sin \delta }=\frac{\left[\cos \alpha \cos \beta +\sin \alpha \sin \beta \right]}{\sin \alpha \sin \beta }\times \frac{\left[\sin \gamma \cos \delta -\cos \gamma \sin \delta \right]}{\sin \gamma \sin \delta } \\ \Rightarrow \left[\cot \alpha \cot \beta -1\right]\left[\cot \delta +\cot \gamma \right]=\left[\cot \alpha \cot \beta +1\right]\left[\cot \delta -\cot \gamma \right] \\ \Rightarrow \cot \alpha \cot \beta cot\delta +\cot \alpha \cot \beta cot\gamma -cot\delta -cot\gamma =\cot \alpha \cot \beta cot\delta -\cot \alpha \cot \beta cot\gamma +cot\delta -cot\gamma \\ \Rightarrow -\cot \delta -\cot \delta =-\cot \alpha \cot \beta \cot \gamma -\cot \alpha \cot \beta \cot \gamma \\ \Rightarrow -2\cot \delta =-2\cot \alpha \cot \beta \cot \gamma \\ \Rightarrow \cot \alpha \cot \beta \cot \gamma =\cot \delta \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$