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Question: If \(A + B + C = 180^{o},\) then the value of \((\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B)...

If A+B+C=180o,A + B + C = 180^{o}, then the value of

(cot⁡B+cot⁡C)(cot⁡C+cot⁡A)(cot⁡A+cot⁡B)(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B) will be

A

sec⁡Asec⁡Bsec⁡C\sec A\sec B\sec C

B

cosecAcosecBcosecCc\text{osec}Aco\text{sec}B\text{cosec}C

C

tan⁡Atan⁡Btan⁡C\tan A\tan B\tan C

D

1

Answer

cosecAcosecBcosecCc\text{osec}Aco\text{sec}B\text{cosec}C

Explanation

Solution

cot⁡B+cot⁡C=sin⁡Ccos⁡B+sin⁡Bcos⁡Csin⁡B.sin⁡C\cot B + \cot C = \frac{\sin C\cos B + \sin B\cos C}{\sin B.\sin C}

=sin⁡(B+C)sin⁡B.sin⁡C=sin⁡(180o−A)sin⁡B.sin⁡C=sin⁡Asin⁡B.sin⁡C\frac{\sin(B + C)}{\sin B.\sin C} = \frac{\sin(180^{o} - A)}{\sin B.\sin C} = \frac{\sin A}{\sin B.\sin C}Similarly, cot⁡C+cot⁡A=sin⁡Bsin⁡C.sin⁡A\cot C + \cot A = \frac{\sin B}{\sin C.\sin A} and cot⁡A+cot⁡B=sin⁡Csin⁡Asin⁡B\cot A + \cot B = \frac{\sin C}{\sin A\sin B}

Therefore, (cot⁡B+cot⁡C)(cot⁡C+cot⁡A)(cot⁡A+cot⁡B)(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B)

= sin⁡Asin⁡B.sin⁡C.sin⁡Bsin⁡C.sin⁡A.sin⁡Csin⁡Asin⁡B=cosecA.cosecB.cosecC\frac{\sin A}{\sin B.\sin C}.\frac{\sin B}{\sin C.\sin A}.\frac{\sin C}{\sin A\sin B} = \text{cosec}A.\text{cosec}B.\text{cosec}C.