sin2A  sinC  sinB

sinC    sin2B  sinA

sinB   sinA      sin2C

ye ek determinant mein hain....actly mjhe formula editor use karna nahi aaya.......

so the question is if A,B and C are the angles of a triangle then prove that determinant is 0

expand sin 2a=2sin acos a

perform C₂=C₂-C₃ and put  sin(B+C)/2=cosA/2 and solve u will get the ans

hope u got it..

hey listen i m giving u a shortcut method also

put sin C=sin (180-(B+A)) =sin (B+A) and similarly write sin B also.

now write the determinant as the product of 2 determinants as

sin 2A   sin (A+B)   sin(A+C)

sin C     sin 2B        sin A

sin B      si

now write the determinant as the product of 2 determinants as

sin A    cos A   0                            cos A    cos B    cos C 

.....        .....       ...                *           sin A      sin B      sin C

....       ....          ....                               0             0             0

so the value comes out to b 0

hope u got it......