If \[ \cos(\alpha+\beta)\sin(\gamma+\delta) = \cos(\alpha-\beta)\sin(\gamma-\delta) \] prove that \[ \cot\alpha\cot\beta\cot\gamma=\cot\delta \]
Solution
Given:
\[ \cos(\alpha+\beta)\sin(\gamma+\delta) = \cos(\alpha-\beta)\sin(\gamma-\delta) \]Use identities:
\[ \cos X\sin Y = \frac12[\sin(X+Y)+\sin(Y-X)] \]
\[
\frac12[
\sin(\alpha+\beta+\gamma+\delta)
+
\sin(\gamma+\delta-\alpha-\beta)
]
\]
\[
=
\frac12[
\sin(\alpha-\beta+\gamma-\delta)
+
\sin(\gamma-\delta-\alpha+\beta)
]
\]
Multiply by 2:
\[ \sin(\alpha+\beta+\gamma+\delta) + \sin(\gamma+\delta-\alpha-\beta) \] \[ = \sin(\alpha-\beta+\gamma-\delta) + \sin(\gamma-\delta-\alpha+\beta) \]Use identity:
\[ \sin C+\sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \]
\[
2\sin(\gamma+\delta)\cos(\alpha+\beta)
=
2\sin(\gamma-\delta)\cos(\alpha-\beta)
\]
Expand using compound angle formulas:
\[ (\sin\gamma\cos\delta+\cos\gamma\sin\delta) (\cos\alpha\cos\beta-\sin\alpha\sin\beta) \] \[ = (\sin\gamma\cos\delta-\cos\gamma\sin\delta) (\cos\alpha\cos\beta+\sin\alpha\sin\beta) \]Expand and simplify:
\[ 2\cos\alpha\cos\beta\cos\gamma\sin\delta = 2\sin\alpha\sin\beta\sin\gamma\cos\delta \]
\[
\cos\alpha\cos\beta\cos\gamma\sin\delta
=
\sin\alpha\sin\beta\sin\gamma\cos\delta
\]
Divide both sides by
\[ \sin\alpha\sin\beta\sin\gamma\sin\delta \]
\[
\cot\alpha\cot\beta\cot\gamma
=
\cot\delta
\]
Hence Proved.