Question

If \[\dfrac{x}{\cos \alpha }=\dfrac{y}{\cos \left( \alpha -\dfrac{2\pi }{3} \right)}=\dfrac{z}{\cos \left( \alpha +\dfrac{2\pi }{3} \right)}\], then \[x+y+z\] is equal to
A.1
B.0
C.-1
D. none of these

Answer 1

Hint: Here, this problem is solved with the trigonometric formulas. Firstly, we consider the above equation is equal to arbitrary constant ‘t’, then equalizing all, and using this trigonometric formula \[\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\] , then we evaluate the \[x+y+z\]and finally we find the correct option.

Complete step by step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric functions.
Let us assume the given expression is equal to arbitrary constant ’t’
\[\dfrac{x}{\cos \alpha }=\dfrac{y}{\cos \left( \alpha -\dfrac{2\pi }{3} \right)}=\dfrac{z}{\cos \left( \alpha +\dfrac{2\pi }{3} \right)}=t\]
Here, we are equalizing all the individual terms to arbitrary constant ’t’, we are multiplied the denominator part of the x, y and z terms with constants ’t’, then we get
\[x=t\cos \alpha \]
\[y=t\cos \left( \alpha -\dfrac{2\pi }{3} \right)\]
\[z=t\cos \left( \alpha +\dfrac{2\pi }{3} \right)\]
Now, we are solving for the expression \[x+y+z\], substituting the x, y and z values in the \[x+y+z\]
\[\Rightarrow x+y+z=t\cos \alpha +t\cos \left( \alpha -\dfrac{2\pi }{3} \right)+t\cos \left( \alpha +\dfrac{2\pi }{3} \right)\]
Here, we are taking the ‘t’ common from Right Hand Side, then we obtain
\[\Rightarrow x+y+z=t\left( \cos \alpha +\cos \left( \alpha -\dfrac{2\pi }{3} \right)+\cos \left( \alpha +\dfrac{2\pi }{3} \right) \right)\]
From the above equation, we have to apply the basic formula to evaluate,
As we now the cosine sum identity i.e., \[\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\]
Evaluating the expression, on basis of the cosine sum identity We get
Here, \[C=\alpha \]and \[D=\left( \dfrac{2\pi }{3} \right)\]
Substituting the Cand D values in the expression we get
\[\Rightarrow t\left( \cos \alpha +2\cos \alpha \cos \left( \dfrac{-2\pi }{3} \right) \right)\]
We know that, \[\cos \left( \dfrac{-2\pi }{3} \right)=\dfrac{-1}{2}\]
On substituting,
\[\Rightarrow t\left( \cos \alpha +2\cos \alpha \left( \dfrac{-1}{2} \right) \right)\]
\[\Rightarrow t\left( \cos \alpha -\cos \alpha \right)\]
\[=0\]
Therefore, \[x+y+z=0\]

So, the correct answer is “Option B”.

Note: When solving trigonometry-based questions, we have to know the definitions of ratios and always remember the standard angles and formulas are useful for solving certain integration problems where a cosine and sum identity may make things much simpler to solve. Thus, in math as well as physics, these formulae are useful to derive many important identities.