Question #5dcd0

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Answer 1 · 2017-11-30T06:50:25

The relation is #A=pi/4#

The first relation is

#(tanA-tanB)/(1+tanAtanB)=tan(A-B)#

You can easily prove this identity using

#sin(A-B)=sinAcosB-sinBcosA#

and

#cos(A-B)=cosAcosB+sinAsinB#

#tan(A-B)=sin(A-B)/cos(A-B)#

#=(sinAcosB-sinBcosA)/(cosAcosB+sinAsinB)#

#=((sinAcosB)/(cosAcosB)-(sinBcosA)/(cosAcosB))/((cosAcosB)/(cosAcosB)+(sinAsinB)/(cosAcosB))#

#=(tanA-tanB)/(1+tanAtanB)#

Let #A=pi/4#

#tan(pi/4-B)=(tan(pi/4)-tanB)/(1+tan(pi/4)tanB)#

#=(1-tanB)/(1+tanB)#