In triangle ABC,(cot A+cot B)(cot B+cot C)(co | vedclass

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Question

(C) Given the expression $(\cot A+\cot B)(\cot B+\cot C)(\cot C+\cot A)$.Using $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$,we have:$= \left(\frac{

Vedclass · Accepted Answer

$\operatorname{cosec} A \operatorname{cosec} B \operatorname{cosec} C$

Vedclass · Answer

(C) Given the expression $(\cot A+\cot B)(\cot B+\cot C)(\cot C+\cot A)$.Using $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$,we have:$= \left(\frac{\cos A}{\sin A} + \frac{\cos B}{\sin B}ight) \left(\frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}ight) \left(\frac{\cos C}{\sin C} + \frac{\cos A}{\sin A}ight)$$= \left(\frac{\sin(A+B)}{\sin A \sin B}ight) \left(\frac{\sin(B+C)}{\sin B \sin C}ight) \left(\frac{\sin(C+A)}{\sin C \sin A}ight)$Since $A+B+C = \pi$,we have $\sin(A+B) = \sin(\pi-C) = \sin C$,$\sin(B+C) = \sin A$,and $\sin(C+A) = \sin B$.Substituting these values:$= \left(\frac{\sin C}{\sin A \sin B}ight) \left(\frac{\sin A}{\sin B \sin C}ight) \left(\frac{\sin B}{\sin C \sin A}ight)$$= \frac{\sin A \sin B \sin C}{(\sin A \sin B \sin C)^2} = \frac{1}{\sin A \sin B \sin C}$$= \operatorname{cosec} A \operatorname{cosec} B \operatorname{cosec} C$.

In $\triangle ABC$,$(\cot A+\cot B)(\cot B+\cot C)(\cot C+\cot A) =$

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