If (B+C-A), (C+A-B), (A+B-C) are in A.P., than , , are in

MCQ

If $\sin(\text{B+C}-\text{A}),\sin(\text{C+A}-\text{B}),\sin(\text{A+B}-\text{C})$ are in $A.P.,$ than $\cot\text{A},\cot\text{B},\cot\text{C}$ are in

A
$\text{GP}$
✓
$\text{HP}$
C
$\text{AP}$
D
None of these

✓

Answer

Correct option: B.

$\text{HP}$

Given:
$\sin(\text{B+C}-\text{A}),\sin(\text{C+A}-\text{B}),$ and $\sin(\text{A+B}-\text{C})$ are in $A.P$.
$\Rightarrow\ \sin(\text{C+A}-\text{B})-\sin(\text{B+C}-\text{A})$
$\ \ \ =\sin(\text{A+B}-\text{C})-\sin(\text{C+A}-\text{B})$
$\Rightarrow\ 2\sin\Big(\frac{\text{C+A}-\text{B}-\text{B}-\text{C+A}}{2}\Big)\cos\Big(\frac{\text{C+A}-\text{B+B+C}-\text{A}}{2}\Big)$
$\ \ \ =2\sin\Big(\frac{\text{A+B}-\text{C}-\text{C}-\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A+B}-\text{C+C+A}-\text{B}}{2}\Big)$
$\Rightarrow\ \sin(\text{A}-\text{B})\cos\text{C}=\sin(\text{B}-\text{C})\cos\text{A}$
$\Rightarrow\ \sin\text{A}\cos\text{B}\cos\text{C}-\cos\text{A}\sin\text{B}\cos\text{C}$
$\ \ =\sin\text{B}\cos\text{C}\cos\text{A}-\cos\text{B}\sin\text{C}\cos\text{A}$
$\Rightarrow\ 2\sin\text{B}\cos\text{A}\cos\text{C}$
$=\sin\text{A}\cos\text{B}\cos\text{C}+\cos\text{A}\cos\text{B}\sin\text{C}$
Dividing both sides by $\cos\text{A}\cos\text{B}\cos\text{C}:$
$2\tan\text{B}=\tan\text{A}+\tan\text{C}$
$\Rightarrow\ \frac{2}{\cot\text{B}}=\frac{1}{\cot\text{A}}+\frac{1}{\cot\text{C}}$
Hence, $\cot\text{A}, \cot\text{B}$ and $\cot\text{C}$ are in $H.P$.

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