Question

 prove the equations
tan ²Ф+3=3secФ

Answer 1

$tan^2\phi+3=3sec\phi;\ \phi\neq\frac{\pi}{2}+k\pi\ (k\in\mathbb{Z})\\\\\frac{sin^2\phi}{cos^2\phi}+3=\frac{3}{cos\phi}\\\\\frac{sin^2\phi}{cos^2\phi}+\frac{3cos^2\phi}{cos^2\phi}-\frac{3}{cos\phi}=0\\\\\frac{sin^2\phi+3cos^2\phi}{cos^2\phi}-\frac{3cos\phi}{cos^2\phi}=0\\\\\frac{1-cos^2\phi+3cos^2\phi-3cos\phi}{cos^2\phi}=0\\\\\frac{2cos^2\phi-3cos\phi+1}{cos^2\phi}=0\iff2cos^2\phi-3cos\phi+1=0\\\\put\cos\phi=t\in <-1;\ 1>$

$2t^2-3t+1=0\\\\2t^2-2t-t+1=0\\\\2t(t-1)-1(t-1)=0\\\\(t-1)(2t-1)=0\iff t-1=0\ or\ 2t-1=0\\\\t=1\ or\ 2t=1\\\\t=1\ or\ t=\frac{1}{2}\\\\cos\phi=1\ or\ cos\phi=\frac{1}{2}\\\\\phi=\frac{\pi}{2}+2k\pi\notin D\\\\\phi=\frac{\pi}{3}+2k\pi\ or\ \phi=-\frac{\pi}{3}+2k\pi\ (k\in\mathbb{Z})$

This is not a trigonometric identity.