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) tanx+tany1−tanx⋅tany\frac{ tanx+tany}{1-tanx \cdot tany}1−tanx⋅tanytanx+tany =(p-q) q1−tanx+tanycotx+coty\frac{ q}{1-\frac {tanx + tany}{cotx+coty}}1−cotx+cotytanx+tanyq =(p-q)q1−qp\frac{q}{1-\frac{q}{p}}1−pqq =(p-q) qp−qp\frac{q}{\frac{p-q}{p}}pp−qq =(p-q)pqp−q\frac{pq}{p-q}p−qpq =pq
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