Let sin−1a=A sin−1b=B sin−1C=C ∴ sinA=a,sinB=b,sinC=C and A+B+C=π, then sin2A+sin2B+sin2C=4sinAsinBsinC ...(i) ⇒ sinAcosA+sinBcosB+sinCcosC=2sinAsinBsinC ⇒ sinA√1−sin2A+sinB√1−sin2B+sinC√1−sin2C=2sinAsinBsinC ,..(ii) ⇒ a√1−a2+b√1−b2+c√1−c2=2abc while sin−1a+sin−1b+sin−1c=π