Question

If $\sin (y+z-x),\sin (z+x-y),\sin (x+y-z)$ be in A.P., prove that $\tan x,\tan y,\tan z$ are also in A.P.

Answer 1

We know that if $a,b,c$ are in A.P then
$b-c=c-b$
$\therefore \sin (z+x-y)-\sin (y+z-x)=\sin (x+y-z)-\sin (z+x-y)$
$2\cos z\sin (x-y)=2\cos x\sin (y-z)$
Divide both sides by $\cos x\cos y\cos z$
$\tan x-\tan y-\tan y-\tan z$
or $2\tan y=\tan x+\tan z$
$\tan x,\tan y,\tan z$ are also in A.P

Hence proved.