Question

In a ∆ABC, b²+ c² = 1999a², then $\frac { \cot \mathrm { B } + \cot \mathrm { C } } { \cot \mathrm { A } } =$

• A. $\frac { 1 } { 999 }$
• B. $\frac { 1 } { 1999 }$
• C. 999
• D. 1999

Answer 1

Answer: (A)

$\frac { 1 } { 999 }$

Answer 2

$\frac { \cot B + \cot C } { \cot A } = \frac { \frac { \cos B } { \sin B } + \frac { \cos C } { \sin C } } { \cos A / \sin A } = \frac { \sin ( B + C ) \sin A } { \cos A \sin B \sin C }$ $\frac { 2 a ^ { 2 } } { \left( b ^ { 2 } + c ^ { 2 } - a ^ { 2 } \right) }$