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Question: In a \(b ^ { 2 } + c ^ { 2 } = 3 a ^ { 2 }\), then \(\cot B + \cot C - \cot A =\)....

In a b2+c2=3a2b ^ { 2 } + c ^ { 2 } = 3 a ^ { 2 }, then cot⁡B+cot⁡C−cot⁡A=\cot B + \cot C - \cot A =.

A

1

B

ab4Δ\frac { a b } { 4 \Delta }

C

0

D

ac4Δ\frac { a c } { 4 \Delta }

Answer

0

Explanation

Solution

cot⁡B+cot⁡C−cot⁡A=cos⁡Bsin⁡B+cos⁡Csin⁡C−cot⁡A\cot B + \cot C - \cot A = \frac { \cos B } { \sin B } + \frac { \cos C } { \sin C } - \cot A

=sin⁡Ccos⁡B+cos⁡Csin⁡Bsin⁡Bsin⁡C−cot⁡A= \frac { \sin C \cos B + \cos C \sin B } { \sin B \sin C } - \cot A =sin⁡(B+C)sin⁡Bsin⁡C−cos⁡Asin⁡A= \frac { \sin ( B + C ) } { \sin B \sin C } - \frac { \cos A } { \sin A }

=sin⁡2A−sin⁡Bsin⁡Ccos⁡Asin⁡Asin⁡Bsin⁡C=a2−bccos⁡Ak(abc)= \frac { \sin ^ { 2 } A - \sin B \sin C \cos A } { \sin A \sin B \sin C } = \frac { a ^ { 2 } - b c \cos A } { k ( a b c ) }

Sincesin⁡Aa=sin⁡Bb=sin⁡Cc=k\frac { \sin A } { a } = \frac { \sin B } { b } = \frac { \sin C } { c } = k (say)

and cos⁡A=b2+c2−a22bc\cos A = \frac { b ^ { 2 } + c ^ { 2 } - a ^ { 2 } } { 2 b c } =a2−bc(b2+c2−a2)2bc(abc)k= \frac { a ^ { 2 } - b c \frac { \left( b ^ { 2 } + c ^ { 2 } - a ^ { 2 } \right) } { 2 b c } } { ( a b c ) k }

=(a2−a2)abck=0,{= \frac { \left( a ^ { 2 } - a ^ { 2 } \right) } { a b c k } = 0 , \left\{ \right. As b2+c2−a22=3a2−a22=2a22=a2}\left. \frac { b ^ { 2 } + c ^ { 2 } - a ^ { 2 } } { 2 } = \frac { 3 a ^ { 2 } - a ^ { 2 } } { 2 } = \frac { 2 a ^ { 2 } } { 2 } = a ^ { 2 } \right\}