Answer: B
Given: \[ \sin 3\theta \cdot \sec 6\theta = 1 \] So, \[ \sin 3\theta = \frac{1}{\sec 6\theta} \] Which implies: \[ \sin 3\theta = \cos 6\theta \] Now, \[ \sin 3\theta = \sin(90^\circ - 6\theta) \] Therefore, \[ 3\theta = 90^\circ - 6\theta \] Adding both sides: \[ 3\theta + 6\theta = 90^\circ \] \[ 9\theta = 90^\circ \] \[ \theta = 10^\circ \] Then, \[ 6\theta = 60^\circ \] Hence, \[ \tan 6\theta = \tan 60^\circ = \sqrt{3} \]
Given: \[ \sin 3\theta \cdot \sec 6\theta = 1 \] So, \[ \sin 3\theta = \frac{1}{\sec 6\theta} \] Which implies: \[ \sin 3\theta = \cos 6\theta \] Now, \[ \sin 3\theta = \sin(90^\circ - 6\theta) \] Therefore, \[ 3\theta = 90^\circ - 6\theta \] Adding both sides: \[ 3\theta + 6\theta = 90^\circ \] \[ 9\theta = 90^\circ \] \[ \theta = 10^\circ \] Then, \[ 6\theta = 60^\circ \] Hence, \[ \tan 6\theta = \tan 60^\circ = \sqrt{3} \]