Question

$\sin (y+z-x),\sin (z+x-y),\sin (x+y-z)$ are in A.P., then $\tan x,\tan y,\tan z$ are in

• A. A.G.P.
• B. G.P.
• C. H.P.
• D. None of these
  

Answer 1

Answer: (D)

None of these

$\sin (y+z-x)+\sin (x+y-z)$
$=2\sin \left(y\right)\cos (x+z)$
$=2\sin (z+x-y)$
Hence
$\sin (z+x-y)=\sin \left(y\right)\cos (x+z)$
$\sin (z+x)\cos \left(y\right)-\cos (z+x)\sin y=\sin \left(y\right)\cos (x+z)$
$\sin (z+x)\cos \left(y\right)=2\sin \left(y\right)\cos (x+z)$
Hence
$\tan (x+z)=2\tan y$