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Question: In a triangle ABC, \(\frac { \cos B + \cos C } { 1 - \cos A }\)=...

In a triangle ABC, cos⁡B+cos⁡C1−cos⁡A\frac { \cos B + \cos C } { 1 - \cos A }=

A

b+c1−a\frac { b + c } { 1 - a }

B

bc1−a\frac { \mathrm { bc } } { 1 - \mathrm { a } }

C

b+ca\frac { b + c } { a }

D
Answer

b+ca\frac { b + c } { a }

Explanation

Solution

Q b = c cos A + a cos C , c = a cos B + b cos A

̃ b + c = (b + c) cos A + a (cos B + cos C)

̃ (b + c) (1 – cos A) = a (cos B + cos C)

̃ cos⁡B+cos⁡C1−cos⁡A\frac { \cos B + \cos C } { 1 - \cos A } = b+ca\frac { b + c } { a }