Question

Solve for the angle B tan 2B = cot 2B

Answer 1

$\tan2B=\cot2B$

$\tan2B-\cot2B=0$

$\tan2B-\dfrac1{\tan2B}=0$

$\dfrac{\tan^22B-1}{\tan2B}=0$

Note that we can't have $\tan2B=0$. Meanwhile,

$\tan^22B-1=0\implies\tan^22B=1\implies\tan2B=\pm1$

$\tan2B=\pm1\implies2B=\pm\dfrac\pi4+n\pi$

where $n$ is any integer.

$\implies\boxed{B=\pm\dfrac\pi8+\dfrac{n\pi}2}$