Question

Prove that:

\[\frac{\cos (A + B + C) + \cos ( - A + B + C) + \cos (A - B + C) + \cos (A + B - C)}{\sin (A + B + C) + \sin ( - A + B + C) + \sin (A - B + C) - \sin (A + B - C)} = \cot C\]

Answer 1

Consider LHS: 
\[ \frac{\cos(A + B + C) + \cos( - A + B + C) + \cos(A - B + C) + \cos(A + B - C)}{\sin(A + B + C) + \sin( - A + B + C) + \sin(A - B + C) - \sin(A + B - C)}\]
\[ = \frac{2\cos\left( \frac{A + B + C - A + B + C}{2} \right)\cos\left( \frac{A + B + C + A - B - C}{2} \right) + 2\cos\left( \frac{A - B + C + A + B - C}{2} \right)\cos\left( \frac{A - B + C - A - B + C}{2} \right)}{2\sin\left( \frac{A + B + C - A + B + C}{2} \right)\cos\left( \frac{A + B + C + A - B - C}{2} \right) + 2\sin\left( \frac{A - B + C - A - B + C}{2} \right)\cos\left( \frac{A - B + C + A + B - C}{2} \right)}\]
\[ = \frac{2\cos \left( B + C \right) \cos A + 2\cos A \cos \left( - B + C \right)}{2\sin \left( B + C \right) \cos A + 2\sin \left( - B + C \right) \cos A}\]
\[ = \frac{2\cos A\left[ \cos \left( B + C \right) + \cos\left( - B + C \right) \right]}{2\cos A\left[ \sin\left( B + C \right) + \sin\left( - B + C \right) \right]}\]
\[ = \frac{\cos \left( B + C \right) + \cos \left( - B + C \right)}{\sin\left( B + C \right) + \sin \left( - B + C \right)}\]
\[ = \frac{2\cos \left( \frac{B + C - B + C}{2} \right) \cos \left( \frac{B + C + B - C}{2} \right)}{2\sin\left( \frac{B + C - B + C}{2} \right) \cos \left( \frac{B + C + B - C}{2} \right)}\]
\[ = \frac{\cos C \cos B}{\sin C \cos B}\]
\[ = \cot C\]
 = RHS
Hence, LHS = RHS.