What is the general solution of the equation \( \cot\theta + \tan\theta = 2 \)?
\( \theta = n\pi/2 + \pi/4 \)
\( \theta = n\pi/2 + \pi/6 \)
\( \theta = n\pi + \pi/4 \)
\( \theta = n\pi/2 + \pi/8 \)
Step 1: Write \( \cot\theta + \tan\theta \) in terms of \(\sin\theta\) and \(\cos\theta\). \[ \cot\theta + \tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta \cos\theta} = \frac{1}{\sin\theta \cos\theta} \] Given that: \[ \frac{1}{\sin\theta \cos\theta} = 2 \] Step 2: Simplify. \[ \sin\theta \cos\theta = \frac{1}{2} \] We know that \( \sin 2\theta = 2 \sin\theta \cos\theta \), so: \[ \sin 2\theta = 2 \times \frac{1}{2} = 1 \] Step 3: Solve for \(2\theta\): \[ 2\theta = \frac{\pi}{2} + 2n\pi \] Therefore, \[ \theta = \frac{\pi}{4} + n\pi \] Final Answer: \[ \boxed{\theta = n\pi + \frac{\pi}{4}} \] Correct Option: (c)
The obtuse angle between lines \(2y = x + 1\) and \(y = 3x + 2\) is:
Let the line $\frac{x}{4} + \frac{y}{2} = 1$ meet the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line $x + y = 1$. Let the point P lie on the line $x + y = 1$ such that $\Delta ABP$ is an isosceles triangle with $AP = BP$. Then the distance between M' and P is:
Let \( F_1, F_2 \) \(\text{ be the foci of the hyperbola}\) \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, a > 0, \, b > 0, \] and let \( O \) be the origin. Let \( M \) be an arbitrary point on curve \( C \) and above the X-axis and \( H \) be a point on \( MF_1 \) such that \( MF_2 \perp F_1 F_2, \, M F_1 \perp OH, \, |OH| = \lambda |O F_2| \) with \( \lambda \in (2/5, 3/5) \), then the range of the eccentricity \( e \) is in: