prove that: (A+B+C)+ (-A+B+C)+ (A-B+C)+ (A+B-C) (A+B+C)+ (-A+B+C)+ (A-B+C)- (A+B-C) =

We have, LHS= (A+B+C)+ (-A+B+C)+ (A-B+C)+ (A+B-C) (A+B+C)+ (-A+B+C)+ (A-B+C)- (A+B-C) = 2 A+B+C-A+B+C 2 A+B+C+A-B-C 2 +2 A-B+C+A+B-C 2 A-B+C-A-B+C 2 2 A+B+C-A+B+C 2 A+B+C+A-B-C 2 +2 A-B+C-A-B+C 2 A-B+C+A+B-C 2 = 2 (B+C) +2 (C-B) 2 (B+C) +2 (C-B) = 2 [ (B+C)+ (C-B)] 2 [ (B+C)+ (C-B)] = (B+C)+ (C-B) (B+C)+ (C-B) = 2 B+C+C-B 2 B+C-C+B 2 2 B+C+C-B 2 B+C-C+B 2 = 2 2 = = = RHS (A+B+C)+ (-A+B+C)+ (A-B+C)+ (A+B-C) (A+B+C)+ (-A+B+C)+ (A-B+C)- (A+B-C) = . Hence proved.

prove that: (A+B+C)+ (-A+B+C)+ (A-B+C)+ (A+B-C) (A+B+C)+ (-A+B+C)…

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Question

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

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Answer

We have, $\text{LHS}=\frac{\cos(\text{A+B+C})+\cos(-\text{A+B+C})+\cos(\text{A}-\text{B+C})+\cos(\text{A+B}-\text{C})}{\sin(\text{A+B+C})+\sin(-\text{A+B+C})+\sin(\text{A}-\text{B+C})-\sin(\text{A+B}-\text{C})}$ $=\ \frac{2\cos\Big\{\frac{\text{A+B+C}-\text{A+B+C}}{2}\Big\}\cos\Big\{\frac{\text{A+B+C+A}-\text{B}-\text{C}}{2}\Big\}\\+2\cos\Big\{\frac{\text{A}-\text{B+C+A+B}-\text{C}}{2}\Big\}\cos\Big\{\frac{\text{A}-\text{B+C}-\text{A}-\text{B+C}}{2}\Big\}}{2\sin\Big\{\frac{\text{A+B+C}-\text{A+B+C}}{2}\Big\}\cos\Big\{\frac{\text{A+B+C+A}-\text{B}-\text{C}}{2}\Big\}\\+2\sin\Big\{\frac{\text{A}-\text{B+C}-\text{A}-\text{B+C}}{2}\Big\}\cos\Big\{\frac{\text{A}-\text{B+C+A+B}-\text{C}}{2}\Big\}}$ $=\ \frac{2\cos(\text{B+C})\cos\text{A}+2\cos\text{A}\cos(\text{C}-\text{B})}{2\sin(\text{B+C})\cos\text{A}+2\sin(\text{C}-\text{B})\cos\text{A}}$ $=\ \frac{2\cos\text{A}[\cos(\text{B+C})+\cos(\text{C}-\text{B})]}{2\cos\text{A}[\sin(\text{B+C})+\sin(\text{C}-\text{B})]}$ $=\ \frac{\cos(\text{B+C})+\cos(\text{C}-\text{B})}{\sin(\text{B+C})+\sin(\text{C}-\text{B})}$ $=\ \frac{2\cos\Big\{\frac{\text{B+C+C}-\text{B}}{2}\Big\}\cos\Big\{\frac{\text{B+C}-\text{C+B}}{2}\Big\}}{2\sin\Big\{\frac{\text{B+C+C}-\text{B}}{2}\Big\}\cos\Big\{\frac{\text{B+C}-\text{C+B}}{2}\Big\}}$ $=\ \frac{2\cos\text{C}\cos\text{B}}{2\sin\text{C}\cos\text{B}}$ $=\ \frac{\cos\text{C}}{\sin\text{C}}$ $=\ \cot\text{C}$ $=\ \text{RHS}$ $\therefore\ \frac{\cos(\text{A+B+C})+\cos(-\text{A+B+C})+\cos(\text{A}-\text{B+C})+\cos(\text{A+B}-\text{C})}{\sin(\text{A+B+C})+\sin(-\text{A+B+C})+\sin(\text{A}-\text{B+C})-\sin(\text{A+B}-\text{C})}=\cot\text{C}.$ Hence proved.