Question
If $\sin (A+B)=1$ and $\cos (A-B)=1$, find $A$ and $B$.
✓
Answer
$\sin (A+B)=1=\sin 90^{\circ}$
$A+B=90^{\circ} \quad(1)$
$\cos (A-B)=1=\cos 0^{\circ}$
$\Rightarrow A-B=0^{\circ}(2)$
Solving $(1)$ and $(2)$ we get
$A=45^{\circ}, B=45^{\circ}$
$A+B=90^{\circ} \quad(1)$
$\cos (A-B)=1=\cos 0^{\circ}$
$\Rightarrow A-B=0^{\circ}(2)$
Solving $(1)$ and $(2)$ we get
$A=45^{\circ}, B=45^{\circ}$
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