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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Consider the following two statements :-
Statement p :
The value of $sin120^\circ$ can be derived by taking $\theta = 240^\circ$ in the equation $2sin \frac{\theta}{2} = \sqrt{1+sin \theta} - \sqrt{1-sin \theta}$
Statement q :
The angles A, B, C and D of any quadrilateral ABCD satisfy the equation $cos \left( \frac{1}{2}(A+C) \right) + cos \left( \frac{1}{2}(B+D) \right) = 0$
Then the truth values of p and q are respectively :-
