Answer:
B
Explanation:
(To save time, I'm going to use x instead of θ)
So we have the expression:
![(\sin(x)-\cos(x))^2+(\sin(x)+\cos(x))^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/ir3ehcklv4ou5ae0xnssv3t2rbmiw6egdj.png)
First, expand these binomials:
![=(\sin^2(x)-2\sin(x)\cos(x)+\cos^2(x))+(\sin^2(x)+2\sin(x)\cos(x)+\cos^2(x))](https://img.qammunity.org/2021/formulas/mathematics/high-school/gxec6ekh3wvz09i8l4glfubrhg489y4k35.png)
Combine like terms:
![=\sin^2(x)+\sin^2(x)+\cos^2(x)+\cos^2(x)-2\sin(x)\cos(x)+2\sin(x)\cos(x)\\=2\sin^2(x)+2\cos^2(x)](https://img.qammunity.org/2021/formulas/mathematics/high-school/dyxb7q4z9tp31htvj6z2njd3ih4ekseoqe.png)
Factor out a 2:
![=2(\sin^2(x)+\cos^2(x))](https://img.qammunity.org/2021/formulas/mathematics/high-school/kxopxd9batn6rr8g92u210nwifx5youjj1.png)
The expression inside the parentheses is the Pythagorean Identity:
![\sin^2(x)+\cos^2(x)=1](https://img.qammunity.org/2021/formulas/mathematics/high-school/lqb6e0v8ywqsg5wgi5mwhn8cgicmw36fbr.png)
Substitute:
![=2(1)\\=2](https://img.qammunity.org/2021/formulas/mathematics/high-school/kbmpmmjg6eafdx586nd7b2qxbd6gh9r6mt.png)
The answer is B.