Prove that:

Solution:

(i) Goncidor LHS.

$\frac{\cos (A+B+C)+\cos (-A+B+C)+\cos (A-B+C)+\cos (A+B-C)}{\sin (A+B+C)+\sin (-A+B+C)+\sin (A-B+C)-\sin (A+B-C)}$

$=\frac{2 \cos \left(\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}-\mathrm{A}+\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{A}-\mathrm{B}-\mathrm{C}}{2}\right)+2 \cos \left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{C}+\mathrm{A}+\mathrm{B}-\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{C}-\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}\right)}{2 \sin \left(\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}-\mathrm{A}+\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{A}-\mathrm{B}-\mathrm{C}}{2}\right)+2 \sin \left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{C}-\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{C}+\mathrm{A}+\mathrm{B}-\mathrm{C}}{2}\right)}$

$=\frac{2 \cos (\mathrm{B}+\mathrm{C}) \cos A+2 \cos A \cos (-B+C)}{2 \sin (\mathrm{B}+\mathrm{C}) \cos A+2 \sin (-B+C) \cos A}$

$=\frac{2 \cos A[\cos (\mathrm{B}+\mathrm{C})+\cos (-B+C)]}{2 \cos A[\sin (\mathrm{B}+\mathrm{C})+\sin (-B+C)]}$

$=\frac{\cos (\mathrm{B}+\mathrm{C})+\cos (-B+C)}{\sin (\mathrm{B}+\mathrm{C})+\sin (-B+C)}$

$=\frac{2 \cos \left(\frac{B+C-B+C}{2}\right) \cos \left(\frac{B+C+B-C}{2}\right)}{2 \sin \left(\frac{B+C-B+C}{2}\right) \cos \left(\frac{B+C+B-C}{2}\right)}$

$=\frac{\cos C \cos B}{\sin C \cos B}$

$=\cot C$

= RHS

Hence, LHS = RHS

(ii) Goncidor LHS.

$\sin (B-C) \cos (A-D)+\sin (C-A) \cos (B-D)+\sin (A-B) \cos (C-D)$

Multiplying by 2 :

$2 \sin (B-C) \cos (A-D)+2 \sin (C-A) \cos (B-D)+2 \sin (A-B) \cos (C-D)$

$=\sin (B-C+A-D)+\sin (B-C-A+D)+\sin (C-A+B-D)+\sin (C-A-B+D)+\sin$$(A-B+C-D)+\sin (A-B-C+D)$

$=\sin \{-(C+D-A-B)\}+\sin \{-(A+C-B-D)\}+\sin \{-(A+D-B-C)\}+\sin (C-A-B+D)$$+\sin (A-B+C-D)+\sin (A-B-C+D)$

$=-\sin (C+D-A-B)-\sin (A+C-B-D)-\sin (A+D-B-C)+\sin (C-A-B+D)+\sin (A-B+C-D)$$+\sin (A-B-C+D)$

= 0

= RHS

Hence, LHS = RHS