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