In a triangle ABC `sin (A/2) sin (B/2) sin (C/2) <= 1/8`

In a triange ABC, if sin(A/2) sin (B/2) sin(C/2) = 1/8 prove that the triangle is equilateral.

Statement I If in a triangle ABC sin ^(2) A+sin ^(2)B+sin ^(2)C=2, then one of the angle must be 90^(@). Statement II In any triangles ABC cos 2A+ cos 2B+cos 2C=-1-4 cos A cos B cos C

Ina triangle ABC (sin A) / (2) (sin B) / (2) (sin C) / (2) <= (1) / (8)

Prove that in a triangle ABC , sin^(2)A - sin^(2)B + sin^(2)C = 2sin A *cos B *sin C .

If A, B, C are angle of a triangle ABC, then the value of the determinant |(sin (A/2), sin (B/2), sin (C/2)), (sin(A+B+C), sin(B/2), cos(A/2)), (cos((A+B+C)/2), tan(A+B+C), sin (C/2))| is less than or equal to

In any triangle ABC,sin^(2)A-sin^(2)B+sin^(2)C is always equal to (A) 2sin A sin B cos C(B)2sin A cos B sin C(C)2sin A cos B cos C(D)2sin A sin B sin C

Statement-1: In a triangle ABC, if sin^(2)A + sin^(2)B + sin^(2)C = 2 , then one of the angles must be 90 °. Statement-2: In any triangle ABC cos 2A + cos 2B + cos 2C = -1 - 4 cos A cos B cos C