A+B+C=180 and cosA=cosB.cosC. prove by tanB. tanC=2
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a=180-(b+c) so
cosa = - cos(b+c) = - (cosb cosc - sinb sinc) = - cosb cosc + sinb sinc
But cosa =cosb cos c so
- cosb cosc + sinb sinc = cosb cos c and therefore
sinb sinc = 2 cosb cosc
Dividing through by cosb cosc we get
tanb tanc = 2 q.e.d
cosa = - cos(b+c) = - (cosb cosc - sinb sinc) = - cosb cosc + sinb sinc
But cosa =cosb cos c so
- cosb cosc + sinb sinc = cosb cos c and therefore
sinb sinc = 2 cosb cosc
Dividing through by cosb cosc we get
tanb tanc = 2 q.e.d
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Answered by
29
a=180-(b+c) So,
cosA = - cos(B+C) = - (cosB cosC - sinB sinC) = - cosB cosC + sinB sinC
But cosA =cosB cosC so
- cosB cosC + sinB sinC = cosB cosC and therefore
sinB sinC = 2 cosB cosC
Dividing through by cosB cosC we get
tanB tanC = 2
HENCE PROVED
MARK IT AS BRAINLIEST!!!
cosA = - cos(B+C) = - (cosB cosC - sinB sinC) = - cosB cosC + sinB sinC
But cosA =cosB cosC so
- cosB cosC + sinB sinC = cosB cosC and therefore
sinB sinC = 2 cosB cosC
Dividing through by cosB cosC we get
tanB tanC = 2
HENCE PROVED
MARK IT AS BRAINLIEST!!!
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