In a triangle PQR,  ∠R=π/2.  If  tan⁡(P/2) and  tan⁡(Q/2)  are the roots of the equation ax2+bx+c=0(a≠0)then

It is given that tan⁡P2 and tan⁡Q2 are roots ofax2+bc+c=0∴ tan⁡P2+tan⁡Q2=−ba and  tan⁡P2tan⁡Q2=caWe have  R=π2  and  P+Q+R=π⇒P+Q=π2⇒P2+Q2=π4∴ tan⁡P2+Q2=tan⁡π4→tan⁡P/2+tan⁡Q/21−tan⁡P/2tan⁡Q/2=1⇒ −ba1−ca=1 ⇒     1−ca=−ba⇒a−c=−b⇒a+b=c