If # tan (beta/2)=4tan (alpha/2)#,prove that #tan ((beta-alpha)/2)=(3sin alpha)/(5-3cos alpha)#?

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Answer 1 · 2018-02-21T07:32:07

Given # tan (beta/2)=4tan (alpha/2)#

then #tan ((beta-alpha)/2)#

#=(tan (beta/2)-tan(alpha/2))/(1+tan(beta/2)tan(alpha/2))#

#=(4tan (alpha/2)-tan(alpha/2))/(1+4tan(alpha/2)tan(alpha/2))#
#=(3tan (alpha/2))/(1+4tan^2(alpha/2))#

#=((3sin (alpha/2))/cos(alpha/2))/(1+(4sin^2(alpha/2))/cos^2(alpha/2)#

#=(3*2sin (alpha/2)*cos(alpha/2))/(2cos^2(alpha/2)+8sin^2(alpha/2))#
#=(3sin alpha)/(1+cosalpha+4(1-cosalpha))#

#=(3sin alpha)/(5-3cos alpha)=RHS#