Question

If sin (y+z-x), sin (z+x-y) and sin (x+y-z) are in AP, then, tan x, tan y and tan z are in
A. AP
B. GP
C. HP
D. None of these

Answer 1

Hint: Put the sin values in the basic relation of AP which is the difference of two consecutive terms will be the same and apply suitable identities. We will use \[\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\times \cos \left( \dfrac{A+B}{2} \right)\] and then convert the sin and cos values in the form of tan values using another identity given by \[\tan C-\tan D=\dfrac{\sin \left( C+D \right)}{\cos C\times \cos D}\]

Complete step by step answer:
We are given that sin (y+z-x), sin (z+x-y), sin (x+y-z) are in AP or arithmetic progression. We have to find the relation between tan (x), tan (y) and tan (z).
We know that if a, b, c are in AP then, the difference between each of them is equal, that is,
b - a = c - b
Applying this very relation in given sin values, we get
sin (y+z-x) to be 'a', sin (z+x-y) to be 'b' and sin (x+y-z) to be 'c',
\[\text{sin }\left( \text{z}+\text{x}-\text{y} \right)-s\text{in }\left( \text{y}+\text{z}-\text{x} \right)\text{ }=\text{ sin }\left( \text{x}+\text{y}-\text{z} \right)-\text{sin }\left( \text{z}+\text{x}-\text{y} \right)\]
We know the trigonometric identity given by
\[\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\times \cos \left( \dfrac{A+B}{2} \right)\]
So, we get
\[\begin{align}
  & \text{sin }\left( \text{z}+\text{x}-\text{y} \right)-s\text{in }\left( \text{y}+\text{z}-\text{x} \right)=2\sin \left( x-y \right)\cos z \\
 & \text{sin }\left( \text{x}+\text{y}-\text{z} \right)-\text{sin }\left( \text{z}+\text{x}-\text{y} \right)=2\sin \left( y-z \right)\cos x \\
 & \Rightarrow 2\sin \left( x-y \right)\cos z=2\sin \left( y-z \right)\cos x \\
\end{align}\]
Now, we will divide the equation by \[\cos x\times \cos y\times \cos z\] on both sides to convert the form into tan.
\[\begin{align}
  & \dfrac{2\sin \left( x-y \right)\cos z}{\cos x\times \cos y\times \cos z}=\dfrac{2\sin \left( y-z \right)\cos x}{\cos x\times \cos y\times \cos z} \\
 & \Rightarrow \dfrac{\sin \left( x-y \right)}{\cos x\times \cos y}=\dfrac{\sin \left( y-z \right)}{\cos y\times \cos z} \\
\end{align}\]
We know the trigonometric identity,
\[\tan C-\tan D=\dfrac{\sin \left( C+D \right)}{\cos C\times \cos D}\]
Applying the identities here, we get
\[\tan x-\tan y=\tan y-\tan z\]
The difference between tan x, tan y and tan z is the same. Therefore, these are in AP.
Therefore, option A is correct.

Note: The problem can be done by other methods too, for example, by considering the relation of AP to be 2b = a+c and then proceeding, so, we will have the equation as, \[\text{2sin }\left( \text{z}+\text{x}-\text{y} \right)=s\text{in }\left( \text{y}+\text{z}-\text{x} \right)\text{+sin }\left( \text{x}+\text{y}-\text{z} \right)\] Then using variation identities, this can be converted in terms of tanx, tany and tanz to reach the final conclusion.