Question

If cos x + cos y + cos z = 0 & sin x + sin y + sin z = 0 then find the value of \[\cot \left( {\dfrac{{x + y}}{2}} \right)\]
A) \[\sin z\]
B) \[\cos z\]
C) \[\cot z\]
D) \[2\sin z\]

Answer 1

Hint: Try to get 2 seperate values of \[\cos z\] and \[\sin z\] from the two equations already given, the values will be in terms of \[\cos x,\cos y\& \sin x,\sin y\] then divide \[\cos z\] by \[\sin z\] to get the value of \[cot z\].

Complete Step by Step Solution:
We are given that \[\cos x + \cos y + \cos z = 0\& \sin x + \sin y + \sin z = 0\]
So from \[\cos x + \cos y + \cos z = 0\]
We can get \[\cos x + \cos y = - \cos z\]
Now break it into \[\cos x + \cos y\]
So we will get it as
\[\begin{array}{l}
\cos x + \cos y = - \cos z\\
2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2} = - \cos z...............................................(i)
\end{array}\]
Again from the sine part we will get
\[\begin{array}{l}
\sin x + \sin y = - \sin z\\
2\sin \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2} = - \sin z.................................................(ii)
\end{array}\]
Now if we divide equation (i) by equation (ii)
We will get it as
\[\begin{array}{l}
\therefore \dfrac{{2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}}}{{2\sin \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}}} = \dfrac{{ - \cos z}}{{ - \sin z}}\\
 \Rightarrow \dfrac{{\cos \dfrac{{x + y}}{2}}}{{\sin \dfrac{{x + y}}{2}}} = \dfrac{{\cos z}}{{\sin z}}\\
 \Rightarrow \cot \left( {\dfrac{{x + y}}{2}} \right) = \cot z
\end{array}\]
Therefore option C is the correct option here.

Note: \[\cos (x + y) = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}\& \sin (x + y) = 2\sin \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}\] These formulas are often used so remember them. Also in the solution we are trying to find the value of \[\cot z\] which is equal to \[\dfrac{{\cos z}}{{\sin z}}\] students commonly divide with the multiplicative inverse i.e., \[\dfrac{{\sin z}}{{\cos z}}\] which leads to an incorrect solution.