If tan (A + B) = sqrt{3} and tan (A - B) = fr | vedclass

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Question

(A) Given that $	an (A + B) = \sqrt{3}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $A + B = 60^{\circ} \dots (1)$.Also,$	an (A - B) = \frac{1}{\sqrt

Vedclass · Accepted Answer

$A = 45^{\circ}, B = 15^{\circ}$

Vedclass · Answer

(A) Given that $	an (A + B) = \sqrt{3}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $A + B = 60^{\circ} \dots (1)$.Also,$	an (A - B) = \frac{1}{\sqrt{3}}$.Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $A - B = 30^{\circ} \dots (2)$.Adding equations $(1)$ and $(2)$:$(A + B) + (A - B) = 60^{\circ} + 30^{\circ}$$2A = 90^{\circ}$$A = 45^{\circ}$.Substituting $A = 45^{\circ}$ in equation $(1)$:$45^{\circ} + B = 60^{\circ}$$B = 60^{\circ} - 45^{\circ} = 15^{\circ}$.Thus,$A = 45^{\circ}$ and $B = 15^{\circ}$.

If $\tan (A + B) = \sqrt{3}$ and $\tan (A - B) = \frac{1}{\sqrt{3}}$,where $0^{\circ} < A + B \leq 90^{\circ}$ and $A > B$,find the values of $A$ and $B$.

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