Prove: [(sec^2 x - 6 tan x + 7) / (sec^2 x - 5)] = [(tan x - 4) / (tan x + 2)]

Rhs has tan x terms so in LHS connvert sec^2x to tan ^2 x + 1

(sec^2 x - 6 tan x + 7) = (tan ^2 x+ 1 - 6 tan x + 7) = tan ^2 x - 6 tan x + 8 = (tan x -4)(tan x-2)

sec^2 x- 5 = 1+ tan ^2 x - 5 = tan ^2 x- 4 = (tan x +2)(tan x-2)

by deviding we get the result