We are given that \( \tan \theta = 2 \), and we need to find the value of \( \sec^2 \theta \).
Step 1: Use the trigonometric identity We know the following identity: \[ \sec^2 \theta = 1 + \tan^2 \theta \]
Step 2: Substitute the given value of \( \tan \theta \) Substitute \( \tan \theta = 2 \) into the identity: \[ \sec^2 \theta = 1 + (2)^2 = 1 + 4 = 5 \]
Answer: The value of \( \sec^2 \theta \) is \( 5 \), so the correct answer is option (1).
Let \( M \) and \( m \) respectively be the maximum and the minimum values of \( f(x) = \begin{vmatrix} 1 + \sin^2x & \cos^2x & 4\sin4x \\ \sin^2x & 1 + \cos^2x & 4\sin4x \\ \sin^2x & \cos^2x & 1 + 4\sin4x \end{vmatrix}, \quad x \in \mathbb{R} \) for \( x \in \mathbb{R} \). Then \( M^4 - m^4 \) is equal to:
If \( \alpha>\beta>\gamma>0 \), then the expression \[ \cot^{-1} \beta + \left( \frac{1 + \beta^2}{\alpha - \beta} \right) + \cot^{-1} \gamma + \left( \frac{1 + \gamma^2}{\beta - \gamma} \right) + \cot^{-1} \alpha + \left( \frac{1 + \alpha^2}{\gamma - \alpha} \right) \] is equal to: