Given,

cosθ+sinθ=1
which means, (cosθ+sinθ)² = 1............[1]

But, we know that,

cos²θ+sin²θ=1.......[2]

From [1] and [2], we get,

(cosθ+sinθ)² = cos²θ+sin²θ

⇒ (cos²θ+sin²θ)+2cosθsinθ = cos²θ+sin²θ

⇒ 2cosθsinθ = 0................[3]

Now,

⇒ (cosθ+sinθ)² = 1 From [1]

⇒ (cosθ-sinθ)²+4cosθsinθ = 1

⇒ (cosθ-sinθ)²+2(2cosθsinθ) = 1

⇒ (cosθ-sinθ)²+0 = 1. From [3]

⇒ (cosθ-sinθ)² = 1

⇒ √(cosθ-sinθ)² = √(1)

⇒ (cosθ-sinθ) = ±1

Hence, proved.

Given Equation
To prove that \( \cos \theta - \sin \theta = \pm 1 \) when \( \cos \theta + \sin \theta = 1 \).

Step 1: Square Both Sides
Starting with the equation:
\( \cos \theta + \sin \theta = 1 \)
Square both sides:
\[
(\cos \theta + \sin \theta)^2 = 1^2
\]
This expands to:
\[
\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta = 1
\]

Step 2: Use the Pythagorean Identity
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we substitute:
\[
1 + 2\cos \theta \sin \theta = 1
\]
This simplifies to:
\[
2\cos \theta \sin \theta = 0
\]

Step 3: Solve for \( \cos \theta \) or \( \sin \theta \)
From \( 2\cos \theta \sin \theta = 0 \):
- Either \( \cos \theta = 0 \)
- Or \( \sin \theta = 0 \)

Step 4: Analyze Each Case
- **Case 1:** If \( \cos \theta = 0 \)
Then \( \sin \theta = 1 \) (from the original equation).
Thus, \( \cos \theta - \sin \theta = 0 - 1 = -1 \).
- **Case 2:** If \( \sin \theta = 0 \)
Then \( \cos \theta = 1 \) (from the original equation).
Thus, \( \cos \theta - \sin \theta = 1 - 0 = 1 \).

Conclusion
Combining both cases, we conclude:
\[
\cos \theta - \sin \theta = \pm 1
\]
This proves the statement as required.