Q116 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\text { If } \dfrac{x}{\cos \theta}=\dfrac{y}{\cos \left(\theta+\dfrac{2 \pi}{3}\right)}=\dfrac{z}{\cos \left(\theta-\dfrac{2 \pi}{3}\right)} \text { then } x+y+z \text { is equal to }$
- A$-2$
- B1
- C$-1$
- Correct answerD0
Solution
Let $\dfrac{x}{\cos \theta} = \dfrac{y}{\cos(\theta + \dfrac{2\pi}{3})} = \dfrac{z}{\cos(\theta - \dfrac{2\pi}{3})} = k$.
Then $x = k \cos \theta$, $y = k \cos(\theta + \dfrac{2\pi}{3})$, and $z = k \cos(\theta - \dfrac{2\pi}{3})$.
Summing these values, $x + y + z = k [\cos \theta + \cos(\theta + \dfrac{2\pi}{3}) + \cos(\theta - \dfrac{2\pi}{3})]$.
Using the sum-to-product formula $\cos(A + B) + \cos(A - B) = 2 \cos A \cos B$, we have $\cos(\theta + \dfrac{2\pi}{3}) + \cos(\theta - \dfrac{2\pi}{3}) = 2 \cos \theta \cos(\dfrac{2\pi}{3})$.
Since $\cos(\dfrac{2\pi}{3}) = -\dfrac{1}{2}$, the expression becomes $2 \cos \theta (-\dfrac{1}{2}) = -\cos \theta$.
Substituting this back into the sum, $x + y + z = k [\cos \theta - \cos \theta] = k(0) = 0$.
Answer: 0
Then $x = k \cos \theta$, $y = k \cos(\theta + \dfrac{2\pi}{3})$, and $z = k \cos(\theta - \dfrac{2\pi}{3})$.
Summing these values, $x + y + z = k [\cos \theta + \cos(\theta + \dfrac{2\pi}{3}) + \cos(\theta - \dfrac{2\pi}{3})]$.
Using the sum-to-product formula $\cos(A + B) + \cos(A - B) = 2 \cos A \cos B$, we have $\cos(\theta + \dfrac{2\pi}{3}) + \cos(\theta - \dfrac{2\pi}{3}) = 2 \cos \theta \cos(\dfrac{2\pi}{3})$.
Since $\cos(\dfrac{2\pi}{3}) = -\dfrac{1}{2}$, the expression becomes $2 \cos \theta (-\dfrac{1}{2}) = -\cos \theta$.
Substituting this back into the sum, $x + y + z = k [\cos \theta - \cos \theta] = k(0) = 0$.
Answer: 0
Hence, the correct answer is option: 'D'.