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Verify each identity: (tanx+tany)/(1-tanxtany)=(sinxcosy+cosxsiny)/(cosxcosy-sinxsiny)
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\[tanx=\frac{sinx}{cosx}\]\[tany=\frac{siny}{cosy}\] \[\large{\frac{tanx+tany}{1-tanxtany}=\frac{(\frac{sinx}{cosx})+(\frac{siny}{cosy})}{1-\frac{sinx}{cosx}\frac{siny}{cosy}}}\]now multiply top and bottom by cosxcosy \[\large{=\frac{(\frac{sinx}{\cancel{cosx}})\times \cancel{cosx} cosy +(\frac{siny}{\cancel{cosy}}\times cosx \cancel{cosy})}{(1 \times cosxcosy)-(\frac{sinxsiny}{\cancel{cosxcosy}} \times \cancel{cosxcosy})}}\]
thank you! @lalaly :)
your welcome:D
@alessandro96
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