Question

Prove that : tan (x-y) + tan ( y-z) + tan (z-x) = tan (x-y) tan ( y-z) tan (z-x)

Answer 1

Solution


We know the formula of tan(A + B + C) is:


$\bold{tan (A + B + C) = \dfrac{(tan A + tan B + tan C - tan A*tan B*tan C)}{(1 - tan A tan B- tan B tan C - tan A tan C)}}$


Now, using above formula,

Putting (A = x - y , B = y - z , C = z - x)

we get,

$tan{((x - y) + (y - z) + (z - x))} = \frac{tan(x - y) + tan(y - z) + tan(z - x) - tan(x - y)*tan(y - z)*tan(z - x)}{1 - tan(x - y)*tan(y - z) - tan(y - z)*tan(z - y) - tan(z - y)*tan(x - y)}\\\\\\\implies tan 0=\frac{tan(x - y) + tan(y - z) + tan(z - x) - tan(x - y)*tan(y - z)*tan(z - x)}{1 - tan(x - y)*tan(y - z) - tan(y - z)*tan(z - y) - tan(z - y)*tan(x - y)}$


$[as, we know tan0° = 0]$


$0 = \frac{tan(x - y) + tan(y - z) + tan(z - x) - tan(x - y)*tan(y - z)*tan(z - x)}{1 - tan(x - y)*tan(y - z) - tan(y - z)*tan(z - y) - tan(z - y)*tan(x - y)}\\\\\\ \implies 0=tan(x-y)+tan(y-z)+tan(z-x)-tan(x-y)*tan(y-z)*tan(z-x) \\ \\ \\ \implies tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)*tan(y-z)*tan(z-x)$

$Hence, \boxed{\bold{tan(x - y) tan(y - z) tan(z - x) = tan(x - y) + tan(y - z) + tan(z - x)}}$