Question:
If $\tan (A-B)=\frac{1}{\sqrt{3}}$ and $\tan (A+B)=\sqrt{3}, 0^{\circ}<(A+B)<90^{\circ}$ and $A>B$, then find $A$ and $B$.
Solution:
Here, $\tan (A-B)=\frac{1}{\sqrt{3}}$
$\Rightarrow \tan (A-B)=\tan 30^{\circ} \quad\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]$
⇒ A − B = 30o ...(i)
Also, $\tan (A+B)=\sqrt{3}$
$\Rightarrow \tan (A+B)=\tan 60^{\circ} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right]$
⇒ A + B = 60o ...(ii)
Solving (i) and (ii), we get:
A = 45o and B = 15o
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