If a = sin(x - y) sin(z – u), b = sin(y...

If `a = sin(x - y) sin(z – u), b = sin(y - z) sin(x - u) and c = sin(z - x) sin(y -u), (a != b != c)`, then whichof following(s) are true ?

Similar Questions

Explore conceptually related problems

Prove that sin (x + y ) sin (x -y) + sin ( y+ z) sin (y - z) + sin (z+x) sin (z-x) = 0.

sin(x+y)sin(x-y)+sin(y+z)sin(y-z)+sin(z+x)sin (z-x)=0

Prove that : sin (y + z-x) + sin (z + x-y) + sin (x +y-z)-sin (x+y+z) = 4 sin x sin y sin z .

Show that sin (x-y) + sin (y-z) + sin (z-x) + 4sin((x-y)/(2)) sin( (y-z)/(2)) sin((z-x)/(2)) =0

Prove that sinx siny sin(x-y)+sinysinzsin(y-z) +sinz sinx sin(z-x)+sin(x-y) sin(y -z) sin (z-x) = 0

If x + y = pi + z, then prove that sin ^ (2) x + sin ^ (2) y-sin ^ (2) z = 2sin x sin y cos z

Prove that sin x* sin y*sin(x - y) + sin y *sin z*sin(y- z) + sin z *sin x sin(z - x) + sin(x - y) *sin(y - z)*sin(z -x) = 0 .

Prove that : sin x sin y sin (x - y) + sin y sin z sin (y- z)+ sin z sin x sin (z - x) + sin (x - y) sin (y-z)sin (z- x) = 0 .

What is the value of [sin (y – z) + sin (y + z) + 2 sin y]/[sin (x – z) + sin (x + z) + 2 sin x]?