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

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Question

(b) $\cot B + \cot C = \frac{{\sin C\,\cos B + \sin B\,\cos C}}{{\sin B\,\sin C}}$

$ = \frac{{\sin (B + C)}}{{\sin B\,\sin C}}$

$ = \frac{{\sin

Vedclass · Accepted Answer

${\rm{cosec}}\,A\,{\rm{cosec}}\,B\,{\rm{cosec}}\,C$

Vedclass · Answer

(b) $\cot B + \cot C = \frac{{\sin C\,\cos B + \sin B\,\cos C}}{{\sin B\,\sin C}}$

$ = \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, $\cot C + \cot A = \frac{{\sin B}}{{\sin C\sin A}}$ 

and $\cot A + \cot B = \frac{{\sin C}}{{\sin A\sin B}}$ 

Therefore, $(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B)$

$ = \frac{{\sin A}}{{\sin B\sin C}}.\frac{{\sin B}}{{\sin C\sin A}}.\frac{{\sin C}}{{\sin A\sin B}}$

$ = \cos {\rm{ec}}A\cos {\rm{ec}}B\cos {\rm{ec}}C.$

If $A + B + C = {180^o},$ then the value of $(\cot B + \cot C)$ $(\cot C + \cot A)\,\,(\cot A + \cot B)$ will be

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