If acos(x+y)=bcos(x-y) then prove (a+b)tanx=(a-b)coty
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acos (x+y)=bcos (x-y)
a(cosx.cosy-sinx.siny)=b ( cosx.cosy+sinx.siny )
(a+b)cosx.cosy =(a-b)sinx.siny
(a+b)sinx/cosx=(a-b)cosy/siny
(a+b)tanx=(a-b)coty
a(cosx.cosy-sinx.siny)=b ( cosx.cosy+sinx.siny )
(a+b)cosx.cosy =(a-b)sinx.siny
(a+b)sinx/cosx=(a-b)cosy/siny
(a+b)tanx=(a-b)coty
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