Question

Prove that:
(i) $\frac{\cos (A+B+C)+\cos (-A+B+C)+\cos (A-B+C)+\cos (A+B-C)}{\sin (A+B+C)+\sin (-A+B+C)+\sin (A-B+C)-\sin (A+B-C)}=\cot C$

(ii) sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\cos (A+B+C)+\cos (-A+B+C)+\cos (A-B+C)+\cos (A+B-C)}{\sin (A+B+C)+\sin (-A+B+C)+\sin (A-B+C)-\sin (A+B-C)} \\ =\frac{2\cos \left({\displaystyle \frac{\mathrm{A}+\mathrm{B}+\mathrm{C}-\mathrm{A}+\mathrm{B}+\mathrm{C}}{2}}\right)\cos \left({\displaystyle \frac{\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{A}-\mathrm{B}-\mathrm{C}}{2}}\right)+2\cos \left({\displaystyle \frac{\mathrm{A}-\mathrm{B}+\mathrm{C}+\mathrm{A}+\mathrm{B}-\mathrm{C}}{2}}\right)\cos \left({\displaystyle \frac{\mathrm{A}-\mathrm{B}+\mathrm{C}-\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}}\right)}{2\sin \left({\displaystyle \frac{\mathrm{A}+\mathrm{B}+\mathrm{C}-\mathrm{A}+\mathrm{B}+\mathrm{C}}{2}}\right)\cos \left({\displaystyle \frac{\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{A}-\mathrm{B}-\mathrm{C}}{2}}\right)+2\sin \left({\displaystyle \frac{\mathrm{A}-\mathrm{B}+\mathrm{C}-\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}}\right)\cos \left({\displaystyle \frac{\mathrm{A}-\mathrm{B}+\mathrm{C}+\mathrm{A}+\mathrm{B}-\mathrm{C}}{2}}\right)} \\ =\frac{2\cos \left(\mathrm{B}+\mathrm{C}\right)\cos A+2\cos A\cos \left(-B+C\right)}{2\sin \left(\mathrm{B}+\mathrm{C}\right)\cos A+2\sin \left(-B+C\right)\cos A} \\ =\frac{2\cos A\left[\cos \left(\mathrm{B}+\mathrm{C}\right)+\cos \left(-B+C\right)\right]}{2\cos A\left[\sin \left(\mathrm{B}+\mathrm{C}\right)+\sin \left(-B+C\right)\right]} \\ =\frac{\cos \left(\mathrm{B}+\mathrm{C}\right)+\cos \left(-B+C\right)}{\sin \left(\mathrm{B}+\mathrm{C}\right)+\sin \left(-B+C\right)} \\ =\frac{2\cos \left({\displaystyle \frac{B+C-B+C}{2}}\right)\cos \left({\displaystyle \frac{B+C+B-C}{2}}\right)}{2\sin \left({\displaystyle \frac{B+C-B+C}{2}}\right)\cos \left({\displaystyle \frac{B+C+B-C}{2}}\right)} \\ =\frac{\cos C\cos B}{\sin C\cos B} \\ =\cot C \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \sin (B-C)\cos (A-D)+\sin (C-A)\cos (B-D)+\sin (A-B)\cos (C-D) \\ \mathrm{Multiplying}\mathrm{by}2: \\ 2\sin (B-C)\cos (A-D)+2\sin (C-A)\cos (B-D)+2\sin (A-B)\cos (C-D) \\ =\sin \left(B-C+A-D\right)+\sin \left(B-C-A+D\right)+\sin \left(C-A+B-D\right)+\sin \left(C-A-B+D\right)+\sin \left(A-B+C-D\right)+\sin \left(A-B-C+D\right) \\ =\sin \left\{-\left(C+D-A-B\right)\right\}+\sin \left\{-\left(A+C-B-D\right)\right\}+\sin \left\{-\left(A+D-B-C\right)\right\}+\sin \left(C-A-B+D\right)+\sin \left(A-B+C-D\right)+\sin \left(A-B-C+D\right) \\ =-\sin \left(C+D-A-B\right)-\sin \left(A+C-B-D\right)-\sin \left(A+D-B-C\right)+\sin \left(C-A-B+D\right)+\sin \left(A-B+C-D\right)+\sin \left(A-B-C+D\right) \\ =0 \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$