Question

3sec^X+8=10sec^X then find the value of tanx

Answer 1

Presumably, "sec^X" is supposed to say $\sec^2x$.

$3\sec^2x+8=10\sec^2x\iff7\sec^2x-8=7(\sec^2x-1)-1=7\tan^2x-1=0$
$\implies\tan^2x=\dfrac17$
$\implies\tan x=\pm\dfrac1{\sqrt7}$