Question #a4b38

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Answer 1 · 2016-12-21T17:09:30

Given

#sin(A+B)/cos(A-B) = (1-m)/(1+m)#

Let #pi/4-A=x and pi/4-B =y#

So #pi/2-(A+B)=(x+y)#

#=>cos(pi/2-(A+B))=cos(x+y)#

#=>sin(A+B)=cos(x+y)#

Again #(A-B)=(y-x)#

#cos(A-B)=cos(y-x)#

Now

#cos(x+y)/cos(y-x) = (1-m)/(1+m)#

#=>cos(x-y)/cos(y+x) = (1+m)/(1-m)#

By componendo and dividendo

#=>(cos(y-x)+cos(y+x))/ (cos(y-x)-cos(y+x))=2/(2m)#

#=>(2cosxcosy)/(2sinxsiny)=1/m#

#=>coty/tanx=1/m#

#=>tanx=mcoty#

#=>tan(pi/4-A)=mcot(pi/4-B)#

Proved