If tanx+tany=q , cotx+coty =p and x+y=c then prove that (p-q)tanc =pq
(p-q)tanc =(p-q)tan(x+y)= (p-q) "\\frac{ tanx+tany}{1-tanx \\cdot tany}" =(p-q) "\\frac{ q}{1-\\frac {tanx + tany}{cotx+coty}}" =(p-q)"\\frac{q}{1-\\frac{q}{p}}" =(p-q) "\\frac{q}{\\frac{p-q}{p}}" =(p-q)"\\frac{pq}{p-q}" =pq
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