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

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

In any triangle ABC b^(2)sin2C+c^(2)sin2B=

In a triangle ABC, 2 ac sin (1/2(A-B + C)) =

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

In a right angled triangle ABC sin^(2)A+sin^(2)B+sin^(2)C=

In triangle ABC if 2sin^(2)C=2+cos2A+cos2B , then prove that triangle is right angled.

In any triangle ABC, show that : 2a sin (B/2) sin (C/2)=(b+c-a) sin (A/2)

In a Delta ABC sin Asin B sin C <= (3sqrt3)/8

In a triangle ABC, if sin A sin(B-C)=sinC sin(A-B) , then prove that cos 2A,cos2B and cos 2C are in AP.

In a triangle ABC, if sin A sin B= (ab)/(c^(2)) , then the triangle is :