Question

If $\sin(y + z - x)$, $\sin(z + x - y)$ and $\sin(x + y - z)$ are in AP, then

• A. $\tan y = \tan x + \tan z$
• B. $\tan y = \tan x - \tan z$
• C. $2\tan y = \tan x + \tan z$
• D. $2\tan y = \tan x - \tan z$

Hint:

Whenever the arguments of sine in an AP take the cyclic configuration $(y+z-x)$, $(z+x-y)$, $(x+y-z)$, their differences reduce to simple linear relationships. Dividing across by the product of cosines will transform a sine AP directly into a standard tangent AP!

Answer 1

The Correct Option is C

Step 1: Translate the AP condition.
If three terms are in AP, the middle term is the average of the outer two, so the common difference is equal on both sides: (2nd term) - (1st term) = (3rd term) - (2nd term).
Step 2: Write the AP condition for the given sines.
$\sin(z+x-y) - \sin(y+z-x) = \sin(x+y-z) - \sin(z+x-y)$.
Step 3: Apply $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$ on both sides.
Left side: half-sum $= z$, half-difference $= x-y$, giving $2\cos z\,\sin(x-y)$. Right side: half-sum $= x$, half-difference $= y-z$, giving $2\cos x\,\sin(y-z)$.
Step 4: Equate and cancel the common factor 2.
$\cos z\,\sin(x-y) = \cos x\,\sin(y-z)$.
Step 5: Expand and divide by $\cos x\cos y\cos z$.
Expanding gives $\cos z(\sin x\cos y - \cos x\sin y) = \cos x(\sin y\cos z - \cos y\sin z)$. Dividing every term by $\cos x\cos y\cos z$ converts each ratio into a tangent: $\tan x - \tan y = \tan y - \tan z$.
Step 6: Rearrange to read off the relation.
Bringing the tangents together gives $\tan x + \tan z = 2\tan y$, which says $\tan x,\ \tan y,\ \tan z$ are themselves in AP.
\[ \boxed{2\tan y = \tan x + \tan z} \]