If $ an(x + y) imes an(x - y) = 1$, then find the value of $ an x$.

Question

If $	an(x + y) 	imes 	an(x - y) = 1$, then find the value of $	an x$.

Accepted Answer

**Solution:**

1. Given the equation:
   $$	an(x + y) 	imes 	an(x - y) = 1$$

2. Let:
   $$A = x + y, \quad B = x - y$$

3. Then the equation becomes:
   $$	an A 	imes 	an B = 1$$

4. Using the tangent addition formula:
   $$	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$$

5. Substitute $	an A 	an B = 1$:
   $$	an(A + B) = \frac{	an A + 	an B}{1 - 1} = \frac{	an A + 	an B}{0}$$

This implies $	an(A + B)$ tends to infinity, so:
   $$A + B = 90^\circ$$

6. Substitute back $A = x + y$ and $B = x - y$:
   $$x + y + x - y = 90^\circ$$
   $$2x = 90^\circ$$
   $$x = 45^\circ$$

7. Therefore:
   $$	an x = 	an 45^\circ = 1$$

**Answer:** $1$

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