`sin^(-1) (ax)/c + sin^(-1) (bx)/c = sin^(-1) x` where `a^2 + b^2 = c^2 and c!=0`

Statement -1: If a^(2)+b^(2)=c^(2),c ne 0 then the non zero solution of the equation sin^(-1)((ax)/(c ))+sin^(-1)((bx)/(c))=sin^(-1)x is pm 1,. Statement-2: sin^(-1)x+sin^(-1)y= sin^(-1)(x+y)

If a^(2) + b^(2) = c^(2), c != 0 , then find the non-zero solution of the equation: sin^(-1).(ax)/(c) + sin^(-1).(bx)/(c) = sin^(-1) x

sin ^ (2) ((A) / (2)) + sin ^ (2) ((B) / (2)) + sin ^ (2) ((C) / (2)) = 1-sin (( A) / (2)) sin ((B) / (2)) sin ((C) / (2))

sin A + sin B + sin C = cos A + cos B + cos C = 0then (A) cos (AB) = - (1) / (2) (B) sin ^ (2) A + sin ^ (2) B + sin ^ (2) C = 0 (C) sin ^ (2) A + sin ^ (2) B + sin ^ (2) C = (3) / (2) (D) cos ^ (2) A + cos ^ (2) B + cos ^ (2) C = (3) / (2)

If sin^(-1)a+sin^(-1)b+sin^(-1)c=pi, then a sqrt(1-a^(2))+b sqrt(1-b^(2))+c sqrt(1-c^(2)) is equal to a+b+c( b )a^(2)b^(2)c^(2)2abc(d)4abc

Let a= ( 2 sin x )/(1 + sin x + cos x ) and b = ( c )/(1+ sin x ) . Then a = b , if c= ?

If sin theta and -cos theta are the roots of the equation ax^(2) - bx - c = 0 , where a, b, and c are the sides of a triangle ABC, then cos B is equal to

If : A+B+C=pi, "then"" "sin ^(2) A +sin^(2)B - sin ^(2)C= A) 2 cos A * cos B * sin C B) 2 cos B * cos C * sin A C) 2 sin A * sin B * cos C D) 2 sin B * sin C * cos A

If A, B and C are the angles of a triangle and |(1,1,1),(1 + sin A,1 + sin B,1 + sin C),(sin A + sin^(2) A,sin B + sin^(2)B,sin C + sin^(2) C)|= 0 , then the triangle ABC is