11. The value of \( \sec 39^{\circ}-\operatorname{cosec} 51^{\circ}= \) \( \quad 0 \)

Yes, your calculation is correct. Let's break it down step by step to understand why: **Given Expression:** \[ \sec 39^{\circ} - \csc 51^{\circ} \] **Step 1: Recall the Definitions** - \(\sec \theta = \frac{1}{\cos \theta}\) - \(\csc \theta = \frac{1}{\sin \theta}\) **Step 2: Substitute the Definitions into the Expression** \[ \sec 39^{\circ} - \csc 51^{\circ} = \frac{1}{\cos 39^{\circ}} - \frac{1}{\sin 51^{\circ}} \] **Step 3: Use Complementary Angle Relationship** Notice that \(51^{\circ}\) and \(39^{\circ}\) are complementary angles because: \[ 51^{\circ} + 39^{\circ} = 90^{\circ} \] And we know that: \[ \sin(90^{\circ} - \theta) = \cos \theta \] So, \[ \sin 51^{\circ} = \cos 39^{\circ} \] **Step 4: Substitute Back into the Expression** \[ \frac{1}{\cos 39^{\circ}} - \frac{1}{\sin 51^{\circ}} = \frac{1}{\cos 39^{\circ}} - \frac{1}{\cos 39^{\circ}} = 0 \] **Conclusion:** \[ \sec 39^{\circ} - \csc 51^{\circ} = 0 \] Your final answer is indeed **0**.