Let `tanA = p, tanB = q and tanC = r`
Then, the given equation becomes,
`(p-q)/(1+pq)+(q-r)/(1+qr)+(r-p)/(1+rp) = 0`
`=>(p-q)(1+qr)(1+rp)+(q-r)(1+pq)(1+rp)+(r-p)(1+qr)(1+pq) = 0`
`=>(p-q)(1+r(p+q)+pqr^2)+(q-r)(1+p(q+r)+qrp^2)+(r-p)(1+q(r+p)+pq^2r) = 0`
`=>[r(p^2-q^2)+(p-q)pqr^2]+[p(q^2-r^2)+(q-r)p^2qr]+[q(r^2-p^2)+(r-p)pq^2r] = 0`
`=>r(p^2-q^2)+p(q^2-r^2)+q(r^2-p^2) = 0`
`=>rp^2-rq^2+pq^2-pr^2+qr^2-qp^2 = 0`
`=>rp^2-rq^2+pq^2-pr^2+qr^2-qp^2+pqr - pqr = 0`
`=>(p-q)(q-r)(r-p) = 0`
`=>p = q or q = r or r = p`
It implies that `ABC` is an isoceles triangle.