Question

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

Answer 1

HELLO DEAR,

we know the formula of tan(A + B + C) is:$\bold{tan (A + B + C) = \frac{(tan A + tan B + tan C - tan A *tan B* tan C)}{(1 - tan A tan B- tan B tan C - tan A tan C)}}$

nos, using above formula,
putting (A = x - y , B = y - z , C = z - x)
we get,
tan{(x - y) + (y - z) + (z - x)} = $\bold{\frac{tan(x - y) + tan(y - z) + tan(z - x) - tan(x - y)*tan(y - z)*tan(z - x)}{1 - tan(x - y)*tan(y - z) - tan(y - z)*tan(z - y) - tan(z - y)*tan(x - y)}}$

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

tan0 = $\bold{\frac{tan(x - y) + tan(y - z) + tan(z - x) - tan(x - y)*tan(y - z)*tan(z - x)}{1 - tan(x - y)*tan(y - z) - tan(y - z)*tan(z - y) - tan(z - y)*tan(x - y)}}$

[as, we know tan0° = 0]

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

0 = tan(x - y) + tan(y - z) + tan(z - x) - tan(x - y)*tan(y - z)*tan(z - x)

on transforming,
we get,
[tan(x - y) tan(y - z) tan(z - x) = tan(x - y) + tan(y - z) + tan(z - x).

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

I HOPE ITS HELP YOU DEAR,
THANKS