Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
In a triangle ABC, prove that acosA+bcosB+ccosC=2atanB×cosC
Dear Aman Yku have typed a wrong RHSIt should be 2a sinB sinC. I have proved here-A/sinA=b/sinB=c/sinC=k (say)∴, a=ksinA, b=ksinB, c=ksinC∴, acosA+bcosB+ccosC=ksinAcosA+ksinBcosB+csinCcosC=k/2(2sinAcosA+2sinBcosB+2sinCcosC)=k/2(sin2A+sin2B+sin2C)=k/2[{2sin(2A+2B)/2cos(2A-2B)/2}+sin2C][∵, sinC+sinD=2sin(C+D)/2cos(C-D)/2]=k/2[2sin(A+B)cos(A-B)+2sinCcosC]=k[sin(π-C)cos(A-B)+sinCcos{π-(A+B)}] [∵, A+B+C=π]=k[sinCcos(A-B)+sinC{-cos(A+B)}]=ksinC[cos(A-B)-cos(A+B)] =ksinC[2sin(A-B+A+B)/2sin(A+B-A+B)/2][∵, cosC-cosD=2sin(C+D)/2sin(D-C)/2]=ksinC(2sinAsinB)=2(ksinA)sinBsinC=2asinBsinC RegardsArun (askIITians forum expert)
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -