Question

If tan(A - 2B) = cot(2A - B) and tan(A + 2B) = cot(2A + B) , show both A and B are multiple of (π/6) and if A is odd multiples then B is even multiples.

Answer 1

If $\tan(A - 2B) = \cot(2A - B)$ and $\tan(A + 2B) = \cot(2A + B)$, show both A and B are multiple of $(\pi/6)$ and if A is odd multiples then B is even multiples.

**Solution:**

$$\cot \alpha = \tan \left(\frac {\pi}{2} - \alpha\right)$$

Hence the first is

$$\tan (A - 2 B) = \tan \left(\frac {\pi}{2} - (2 A - B)\right)$$

$(A - 2B) = \frac{\pi}{2} - (2A - B) + k_1\pi$, where $k_1$ is an integer

And the second is

$$\tan (A + 2 B) = \tan \left(\frac {\pi}{2} - (2 A + B)\right)$$

$(A + 2B) = \frac{\pi}{2} - (2A + B) + k_2\pi$, where $k_2$ is an integer

Solving the system of equations

$$(A - 2 B) = \frac {\pi}{2} - (2 A - B) + k _ {1} \pi$$ $$(A + 2 B) = \frac {\pi}{2} - (2 A + B) + k _ {2} \pi$$

From here

$$3 A - 3 B = \frac {\pi}{2} + k _ {1} \pi$$ $$3 A + 3 B = \frac {\pi}{2} + k _ {2} \pi$$

And finally adding or subtracting

$$6 A = \pi + \left(k _ {1} + k _ {2}\right) \pi$$ $$A = \left(k _ {1} + k _ {2} + 1\right) \frac {\pi}{6}$$ $$6 B = \left(k _ {2} - k _ {1}\right) \pi$$ $$B = \left(k _ {2} - k _ {1}\right) \frac {\pi}{6}$$

Hence A and B are multiple of $\frac{\pi}{6}$. When $k_{1}$ and $k_{2}$ are both odd or even A is odd multiples then B is even multiples, and otherwise A is even multiples then B is odd multiples.