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Question

In a â–³ABC,acotA+bcotB+ccotC is equal to

A
r+R
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B
rR
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C
2(r+R)
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D
2(rR)
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Solution

The correct option is C 2(r+R)
acotA+bcotB+ccotC
=acosAsinA+bcosBsinB+ccosCsinC
Using sine rule asinA=bsinB=csinC=2R we have
=2R(cosA+cosB+cosC)
=2R[2cos(A+B2)cos(AB2)+12sin2C2]
=2R[1+2cos(π2C2)cos(AB2)2sin2C2]
=2R[1+2sin(C2)cos(AB2)2sin2C2]
=2R[1+sin(C2)(cos(AB2)sin(C2))]
=2R[1+sin(C2)(cos(AB2)sin(π2A+B2))]
=2R[1+sin(C2)(cos(AB2)cos(A+B2))]
Using transformation angle formula, we get
=2R[1+sin(C2)2sin(A2)sin(B2)]
=2R[1+4sin(A2)sin(B2)sin(C2)]
=2R(1+rR)
=2(R+r)

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