Question

\(\dfrac{tan\left(x\right)}{1-tan^2\left(x\right)}=\dfrac{1}{2}cot\left(x+\dfrac{\pi}{4}\right)\)

Answer 1

ĐK: \(x\ne\pm\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)

\(\dfrac{tanx}{1-tan^2x}=\dfrac{1}{2}cot\left(x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\dfrac{1}{2}tan2x=\dfrac{1}{2}tan\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow2x=\dfrac{\pi}{4}-x+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\)