Question

If tan (A + B) = $\sqrt{3}$ and tan (A $-$ B) = $\frac{1}{\sqrt{3}}$
0$°$ < A+B$\le$90$°$, A$\ge$B find the value of A and B.

Answer 1

Given:
tan (A + B) = $\sqrt{3}$ and tan (A − B) = $\frac{1}{\sqrt{3}}$
Consider, tan (A + B) = $\sqrt{3}$
$\Rightarrow$ tan (A + B) = tan 60$°$ (From the table)
$\Rightarrow$ A + B = 60$°$ … (1)
Now, consider tan (A − B) = $\frac{1}{\sqrt{3}}$
$\Rightarrow$ tan (A - B) = tan 30$°$ (From table)
$\Rightarrow$ A − B = 30$°$ … (2)
Adding the equations (1) and (2), we get:
A + B + A − B = 60$°$ + 30$°$
$\Rightarrow$ 2A = 90$°$
$\Rightarrow$ A = 45$°$
Putting the value of A in equation (1), we get:
$\Rightarrow$ 45$°$ + B = 60$°$
$\Rightarrow$ B = 60$°$ − 45$°$
$\Rightarrow$ B = 15$°$