If sin (A + B) = 1 and cos (A - B) = frac{sqr | vedclass

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(A) Given $\sin (A + B) = 1$ and $\cos (A - B) = \frac{\sqrt{3}}{2}$.Since $\sin 90^\circ = 1$,we have $A + B = 90^\circ$.Since $\cos 30^\circ = \frac

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$60^\circ, 30^\circ$

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(A) Given $\sin (A + B) = 1$ and $\cos (A - B) = \frac{\sqrt{3}}{2}$.Since $\sin 90^\circ = 1$,we have $A + B = 90^\circ$.Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$,we have $A - B = 30^\circ$.Adding the two equations: $(A + B) + (A - B) = 90^\circ + 30^\circ$ $\Rightarrow 2A = 120^\circ$ $\Rightarrow A = 60^\circ$.Substituting $A = 60^\circ$ in $A + B = 90^\circ$,we get $60^\circ + B = 90^\circ \Rightarrow B = 30^\circ$.Thus,the smallest positive values are $A = 60^\circ$ and $B = 30^\circ$.

If $\sin (A + B) = 1$ and $\cos (A - B) = \frac{\sqrt{3}}{2}$,then the smallest positive values of $A$ and $B$ are

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