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UPSC NDA II 2024 Mathematics Paper

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Question : 32 of 120

Marks: +1, -0

If

\rm \frac{x}{\cos \theta}=\frac{y}{\cos \left(\frac{2\pi}{3}-\theta\right)}=\frac{z}{\cos\left(\frac{2\pi}{3}+\theta\right)}

then what is x + y + z equal to?

$-1$
0
1
3

Solution:

Concept:

Each ratio is set equal to a common constant to express

x, y, z

in terms of

\cos

functions.

Explanation:

Let

\displaystyle \frac{x}{\cos \theta} = \frac{y}{\cos \left(\frac{2\pi}{3} - \theta\right)} = \frac{z}{\cos \left(\frac{2\pi}{3} + \theta\right)} = k

.

Then

x = k \cos \theta

,

y = k \cos \left(\frac{2\pi}{3} - \theta\right)

,

z = k \cos \left(\frac{2\pi}{3} + \theta\right)

.

Now

x + y + z = k \left[ \cos \theta + \cos \left( \frac{2\pi}{3} - \theta \right) + \cos \left( \frac{2\pi}{3} + \theta \right) \right]

.

Using

\cos(A - B) + \cos(A + B) = 2 \cos A \cos B

, we get

\cos \left( \frac{2\pi}{3} - \theta \right) + \cos \left( \frac{2\pi}{3} + \theta \right) = 2 \cos \frac{2\pi}{3} \cos \theta

.

Since

\cos \frac{2\pi}{3} = -\frac{1}{2}

, this equals

2 \cdot \left( -\frac{1}{2} \right) \cos \theta = -\cos \theta

.

Thus the sum inside brackets becomes

\cos \theta - \cos \theta = 0

.

Therefore

x + y + z = k \cdot 0 = 0

.

Answer:

0

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