Saturday, February 11, 2012

if A+B+C = 180° , show that cot A cotB + cotBcotC +cotC cotA=1

if A+B+C = 180° ,
show that

tanA +tanB + tanC = tanAtanBtanC and

cot A cotB + cotBcotC +cotC cotA=1



given A+B+C = 180°

implies A + B = 180°- C

implies tan(A + B) = tan(180°- C)

using trigonometry formulae

tan(A + B) = -tan(C)

therefore

[tanA +tanB] / [1-tanAtanB ] = -tanC

cross multiplying

tanA +tanB = -tanC[1-tanAtanB ]

tanA +tanB = -tanC +tanAtanBtanC

or

tanA +tanB + tanC = tanAtanBtanC

divide each term with tanAtanBtanC we get


cot A cotB + cotBcotC +cotC cotA=1

some other problems
If A+B+C=π,
prove that: sinA+sinB+sinC = cot(A/2).cot(B/2) [sinA+sinB-sinC]

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