Let's solve the problem step-by-step.
We are given:
[tex]\[
\cos A = \frac{1}{2}
\][/tex]
[tex]\[
\cos B = \frac{\sqrt{3}}{2}
\][/tex]
First, we'll calculate [tex]\( \sin A \)[/tex] and [tex]\( \sin B \)[/tex] using the Pythagorean identity:
[tex]\[
\sin^2 A + \cos^2 A = 1
\][/tex]
[tex]\[
\sin^2 B + \cos^2 B = 1
\][/tex]
### Calculating [tex]\( \sin A \)[/tex]:
[tex]\[
\begin{align*}
\cos^2 A &= \left( \frac{1}{2} \right)^2 \\
&= \frac{1}{4}
\end{align*}
\][/tex]
[tex]\[
\begin{align*}
\sin^2 A &= 1 - \cos^2 A \\
&= 1 - \frac{1}{4} \\
&= \frac{3}{4}
\end{align*}
\][/tex]
[tex]\[
\sin A = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}
\][/tex]
### Calculating [tex]\( \sin B \)[/tex]:
[tex]\[
\begin{align*}
\cos^2 B &= \left( \frac{\sqrt{3}}{2} \right)^2 \\
&= \frac{3}{4}
\end{align*}
\][/tex]
[tex]\[
\begin{align*}
\sin^2 B &= 1 - \cos^2 B \\
&= 1 - \frac{3}{4} \\
&= \frac{1}{4}
\end{align*}
\][/tex]
[tex]\[
\sin B = \sqrt{\frac{1}{4}} = \frac{1}{2}
\][/tex]
### Calculating [tex]\( \cos(A + B) \)[/tex]:
Using the angle addition formula for cosine:
[tex]\[
\cos(A + B) = \cos A \cos B - \sin A \sin B
\][/tex]
[tex]\[
\cos A = \frac{1}{2}, \quad \cos B = \frac{\sqrt{3}}{2}, \quad \sin A = \frac{\sqrt{3}}{2}, \quad \sin B = \frac{1}{2}
\][/tex]
[tex]\[
\begin{align*}
\cos(A + B) &= \left( \frac{1}{2} \times \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{3}}{2} \times \frac{1}{2} \right) \\
&= \left( \frac{\sqrt{3}}{4} \right) - \left( \frac{\sqrt{3}}{4} \right) \\
&= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \\
&= 0
\end{align*}
\][/tex]
So the value of [tex]\( \cos(A + B) \)[/tex] is [tex]\( 0 \)[/tex].
The correct answer is (a) 0.
