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If $\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})+\cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})=\frac{1}{2}$ and $0 < \mathrm{B} < \frac{\pi}{2}$, then $\mathrm{B}=$
Options:
Solution:
1938 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{4}$
Given
$$
\begin{aligned}
& \sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B)=\frac{1}{2} \\
& \text { Apply } \cos (x+y)=\cos x \cos y-\sin x \sin y \\
& \Rightarrow \cos (A+B-A+B)=\frac{1}{2} \\
& \cos (2 B)=\frac{1}{2}=\cos \frac{\pi}{3} \\
& 2 B=\frac{\pi}{3} \\
& B=\frac{\pi}{6}
\end{aligned}
$$
Therefore, option (a) is correct.
$$
\begin{aligned}
& \sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B)=\frac{1}{2} \\
& \text { Apply } \cos (x+y)=\cos x \cos y-\sin x \sin y \\
& \Rightarrow \cos (A+B-A+B)=\frac{1}{2} \\
& \cos (2 B)=\frac{1}{2}=\cos \frac{\pi}{3} \\
& 2 B=\frac{\pi}{3} \\
& B=\frac{\pi}{6}
\end{aligned}
$$
Therefore, option (a) is correct.
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