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

\[\cos \left( \alpha + \beta \right) \sin \left( \gamma + \delta \right) = \cos \left( \alpha - \beta \right) \sin \left( \gamma - \delta \right)\]

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

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