Question

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Question:

Prove that sin(B-C)/sin(B+C)=b - cosC - c cos B/ b cos C + c Cos B


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Answer 1

Hello Mate,

Here is your answer,

We Know that,

a = 2RsinA

b = 2RsinB

c = 2RsinC

To Prove:

Sin(B - C) / Sin(B + C) =

Sin(B - C) / Sin(B + C) = b.cosC - c.cosB / b.cosC + c.cosB

2RsinB(cosC) - 2RsinC(cosB)

(divided by)

2RsinB(cosC) + 2RsinC(cosB)

Taking 2R common in numerator and denominator we get,

2R(sinB.cosC + sinC.cosB)

(divided by)

2R(sinB.cosC - sinC.cosB)

We Know sinB.cosC + cosB.sinC =

sin(B + C)

and

We Know sinB.cosC - cosB.sinC =

sin(B - C)

Sin(B - C) / Sin(B + C) =

b.cosC - c.cosB / b.cosC + c.cosB

Hence, Proved.

Hope it helps:)

Answer 2

Answer:

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