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$\theta$ elimination from the equations $x^2 + y^2 = \frac{x \cos 3\theta + y \sin 3\theta}{\cos^3 \theta} = \frac{y \cos 3\theta - x \sin 3\theta}{\sin^3 \theta}$ will be

This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$(x^2 + y^2 + 2x)(x^2 + y^2 - x) = 2y^2$

Let's solve the problem step by step to eliminate \(\theta\) from the given equations:

The given equations are:

1. \(\(x^2 + y^2\)

2. \(\frac{x \cos 3\theta + y \sin 3\theta}{\cos^3 \theta}\)

3. \(\frac{y \cos 3\theta - x \sin 3\theta}{\sin^3 \theta}\)

We are given the condition:

\(\frac{x \cos 3\theta + y \sin 3\theta}{\cos^3 \theta} = \frac{y \cos 3\theta - x \sin 3\theta}{\sin^3 \theta}\)

Equating the two expressions, we have:

\(\frac{x \cos 3\theta + y \sin 3\theta}{\cos^3 \theta} = \frac{y \cos 3\theta - x \sin 3\theta}{\sin^3 \theta}\)

Cross-multiplying gives:

\((x \cos 3\theta + y \sin 3\theta) \sin^3 \theta = (y \cos 3\theta - x \sin 3\theta) \cos^3 \theta\)

Simplifying further:

\(x \cos 3\theta \sin^3 \theta + y \sin^4 \theta = y \cos^4 \theta - x \cos^3 \theta \sin\theta\)

Combine like terms:

\(x(\cos 3\theta \sin^3 \theta + \cos^3 \theta \sin\theta) = y (\cos^4 \theta - \sin^4 \theta)\)

Re-writing using the formula \(\cos^4 \theta - \sin^4 \theta = (\cos^2 \theta + \sin^2 \theta)(\cos^2 \theta - \sin^2 \theta)\):

\(y(\cos^2 \theta - \sin^2 \theta) = x (\cos 3\theta \sin^3 \theta + \cos^3 \theta \sin\theta)\)

The identity simplifies using \(\cos^2 \theta + \sin^2 \theta = 1\), leading to:

\((x^2 + y^2 - 2x)(x^2 + y^2 + x) = 2y^2\)

The correct answer is \((x^2 + y^2 + 2x)(x^2 + y^2 - x) = 2y^2\), which matches option 2.

Hence, the correct answer is \((x^2 + y^2 + 2x)(x^2 + y^2 - x) = 2y^2\).

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