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 \((x^2 + y^2 + 2x)(x^2 + y^2 - x) = 2y^2\).
Similar Questions
- If $\sum_{r=1}^{\infty} \tan^{-1}\left(\frac{1}{2r^2}\right) = a$, then $\tan a$ is equal to
- The true set of values of '$K$' for which $\sin^{-1}\left(\frac{1}{1 + \sin^2 x}\right) = \frac{K\pi}{6}$ may have a solution is
- The general solution of the equation $\sin^{100} x - \cos^{100} x = 1$ is
- If for two real numbers $a, b$ with $|a| \le 1$ and $|b| \le 1$, $\frac{1}{3} + \frac{\sin^{-1} a + \sin^{-1} b}{4} + \frac{\left(\sin^{-1} a + \sin^{-1} b\right)^2}{16} + \frac{\left(\sin^{-1} a + \sin^{-1} b\right)^3}{64} + \dots = \frac{2(8-3\pi)}{3(16+3\pi)}$, then the value of $\sin^{-1}\left(a\sqrt{1-b^2} + b\sqrt{1-a^2}\right)$ is
- Let $g(x) = ax + b$, where $a < 0$ and $g$ is defined from $[1, 3]$ onto $[0, 2]$. Then the value of $\cot \left( \cos^{-1}(|\sin x| + |\cos x|) + \sin^{-1}(-|\cos x| - |\sin x|) \right)$ is equal to
Important Questions from Trigonometry
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
- The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
- Number of solutions of $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi]$ is :
- Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right), \theta \in \mathbf{R}$. Then $\alpha + 2\beta$ is equal to :
- The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{-2} = \frac{z - 2}{3}$. The coordinates of A and B are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively and C is at a distance of 10 units from B. The area (in sq. units) of $\Delta ABC$ is :
