If sin-1a+sin-1b+sin-1c=π, then the value of a1-a2+b1-b2+c1-c2 will be:
2abc
abc
12abc
13abc
Explanation for the correct option.
Step 1: Relation between angle
Given that, sin-1a+sin-1b+sin-1c=π
Put a=sinA,b=sinBandc=sinC
⇒sin-1(sinA)+sin-1(sinB)+sin-1(sinC)=π⇒A+B+C=π
Step 2: Find the value of the expression
So,a1-a2+b1-b2+c1-c2 can be rewritten as:a1-a2+b1-b2+c1-c2=sinAcosA+sinBcosB+sinCcosCMultiply and divide by 2, on RHS we geta1-a2+b1-b2+c1-c2=sin2A+sin2B+sin2C2=2sin(A+B)cos(A-B)+sin2C2=2sinCcos(A-B)+2sinCcosC2…(A+B=π-C)=sinC(cos(A-B)+cos(π-(A+B)))=sinC(cos(A-B)-cos(A+B))=2sinCsinAsinB=2abc
Hence, option A is correct.