Asked by Mansi
2tan^-1(tanx/2×tan(π/4-y/2))=tan^-1(sinx×siny/cosx+cosy)
Answers
Answered by
Steve
You sure there are no typos? I suspect you want
(sinx+siny)/(cosx+cosy)
Anyway, you can get things a bit simpler by taking the tan of both sides, since
tan(arctan(x)) = x
also,
tan(x/2) = (1-cosx)/sinx
(sinx+siny)/(cosx+cosy)
Anyway, you can get things a bit simpler by taking the tan of both sides, since
tan(arctan(x)) = x
also,
tan(x/2) = (1-cosx)/sinx
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