If $ an(A + B) = \sqrt{3}$ and $ an(A - B) = \frac{1}{\sqrt{3}}$ with $0^\circ

Question

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

Accepted Answer

**Solution:**

1. Given:
   - $	an(A + B) = \sqrt{3}$
   - $	an(A - B) = \frac{1}{\sqrt{3}}$

2. Recognize the standard tangent values:
   - $	an 60^\circ = \sqrt{3}$
   - $	an 30^\circ = \frac{1}{\sqrt{3}}$

3. Therefore, we can write:
   - $(A + B) = 60^\circ$  (equation i)
   - $(A - B) = 30^\circ$  (equation ii)

4. Add equations (i) and (ii):
   $$
   (A + B) + (A - B) = 60^\circ + 30^\circ
   $$
   $$
   2A = 90^\circ
   $$
   $$
   A = 45^\circ
   $$

5. Substitute $A = 45^\circ$ into equation (i):
   $$
   45^\circ + B = 60^\circ
   $$
   $$
   B = 60^\circ - 45^\circ = 15^\circ
   $$

6. Since $A > B$, the values are consistent.

**Answer:** $A = 45^\circ$ and $B = 15^\circ$

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