Question : If $\cos(A+B)=0$ and $\sin(A-B)=0$, then what is the value of $\angle A$ and $\angle B$ ? ( $0^\circ < A, B \leq 90 ^\circ$)
Option 1: $20^\circ, 70^\circ$
Option 2: $45^\circ, 45^\circ$
Option 3: $60^\circ, 30^\circ$
Option 4: $15^\circ, 75^\circ$
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Correct Answer: $45^\circ, 45^\circ$
Solution :
Given: $\cos(A+B)=0$ and $\sin(A-B)=0$
The equation $\cos(A+B)=\cos 90^\circ$
$A+B=90^\circ$............(equation 1)
The equation $\sin(A-B)=\sin 0^\circ$
$A-B=0^\circ$............(equation 2)
Solving these two equations, we get:
$\angle A = \angle B = 45^\circ$
Hence, the correct answer is $45^\circ, 45^\circ$.
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