If ` sin^(100) theta - cos^(100) theta =1` , then ` theta ` is

Options

A. `2n pi +(pi)/(3) , n in I `
B. `n pi +(pi)/(2) , n in I `
C. `n pi +(pi)/(4) , n in I `
D. `2n pi -(pi)/(3) , n in I `

Answer

To solve the equation \( \sin^{100} \theta - \cos^{100} \theta = 1 \), we can follow these steps:

### Step 1: Analyze the equation
We start with the equation:
\[ \sin^{100} \theta - \cos^{100} \theta = 1 \]

### Step 2: Understand the implications
For the left-hand side to equal 1, \( \sin^{100} \theta \) must be equal to 1, and \( \cos^{100} \theta \) must be equal to 0. This is because the maximum value of \( \sin^{100} \theta \) is 1 (when \( \sin \theta = 1 \)) and the minimum value of \( \cos^{100} \theta \) is 0 (when \( \cos \theta = 0 \)).

### Step 3: Set up the equations
From the above analysis, we have:
\[ \sin^{100} \theta = 1 \quad \text{and} \quad \cos^{100} \theta = 0 \]

### Step 4: Solve for \( \sin \theta \) and \( \cos \theta \)
1. \( \sin \theta = 1 \)
2. \( \cos \theta = 0 \)

### Step 5: Find the angles
The condition \( \sin \theta = 1 \) occurs at:
\[ \theta = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
The condition \( \cos \theta = 0 \) occurs at:
\[ \theta = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: General solution
Since both conditions lead us to the same set of angles, we can conclude that the general solution for \( \theta \) is:
\[ \theta = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 7: Final answer
Thus, the solution for \( \theta \) is:
\[ \theta = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

---