Prove that: \[ \frac{ \cos(A+B+C)+\cos(-A+B+C)+\cos(A-B+C)+\cos(A+B-C) }{ \sin(A+B+C)+\sin(-A+B+C)+\sin(A-B+C)+\sin(A+B-C) } = \cot C \]
Solution
L.H.S.
\[ = \frac{ \cos(A+B+C)+\cos(-A+B+C) }{ \sin(A+B+C)+\sin(-A+B+C) } \] \[ + \frac{ \cos(A-B+C)+\cos(A+B-C) }{ \sin(A-B+C)+\sin(A+B-C) } \]Use identities:
\[ \cos X+\cos Y = 2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2} \] \[ \sin X+\sin Y = 2\sin\frac{X+Y}{2}\cos\frac{X-Y}{2} \]Simplify numerator:
\[ \cos(A+B+C)+\cos(-A+B+C) = 2\cos(B+C)\cos A \] \[ \cos(A-B+C)+\cos(A+B-C) = 2\cos A\cos(B-C) \]
\[
\text{Numerator}
=
2\cos A[\cos(B+C)+\cos(B-C)]
\]
Again use identity:
\[ \cos X+\cos Y = 2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2} \]
\[
=
2\cos A(2\cos B\cos C)
\]
\[
=
4\cos A\cos B\cos C
\]
Simplify denominator:
\[ \sin(A+B+C)+\sin(-A+B+C) = 2\sin(B+C)\cos A \] \[ \sin(A-B+C)+\sin(A+B-C) = 2\sin A\cos(B-C) \]
\[
\text{Denominator}
=
2\cos A[\sin(B+C)+\sin(B-C)]
\]
Use identity:
\[ \sin X+\sin Y = 2\sin\frac{X+Y}{2}\cos\frac{X-Y}{2} \]
\[
=
2\cos A(2\sin B\cos C)
\]
\[
=
4\cos A\cos B\sin C
\]
\[
=
\frac{
4\cos A\cos B\cos C
}{
4\cos A\cos B\sin C
}
\]
\[
=
\frac{\cos C}{\sin C}
\]
Use identity:
\[ \cot C=\frac{\cos C}{\sin C} \]
\[
=
\cot C
\]
Hence Proved.