Inconsistency with this equation

Sep 2019
2
0
Avalon
Screenshot_20200201-085932_1.png

The question is:-
(secx-1)=(sqrt(2)-1)(tanx)
as shown above in the first line. The task is to find the general solution.
When I substitute secx as sqrt(1+tan^2x) and square both the sides, I get x=0 or pi/4; that is, tanx=0 or tanx=1
But when I substitute tanx as sqrt(sec^2x-1) and square both sides, I don't get the same result.
Can someone point out where am I wrong? I solved it both by myself and by using this calculator.
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
$\sec{x}-1=(\sqrt{2}-1)\tan{x}$

$(\sec{x}-1)^2 = (\sqrt{2}-1)^2\tan^2{x}$

$(\sec{x}-1)^2 = (3-2\sqrt{2})(\sec^2{x}-1)$

$(\sec{x}-1)^2 - (3-2\sqrt{2})(\sec^2{x}-1) = 0$

$(\sec{x}-1)\left[(\sec{x}-1) - (3-2\sqrt{2})(\sec{x}+1)\right] = 0$

first factor ... $\sec{x} - 1 = 0 \implies x = 0$

second factor ... $(\sec{x}-1) - (3-2\sqrt{2})(\sec{x}+1) = 0$

$\sec{x}-1 = 3\sec{x}-2\sqrt{2} \sec{x} + 3 - 2\sqrt{2}$

$0 = (2-2\sqrt{2})\sec{x} + 4 - 2\sqrt{2}$

$\sec{x} = \dfrac{2(\sqrt{2}-2)}{2(1-\sqrt{2})}$

$\sec{x} = \dfrac{\sqrt{2}-2}{1-\sqrt{2}} = \dfrac{\sqrt{2}(1-\sqrt{2})}{1-\sqrt{2}} = \sqrt{2} \implies x = \dfrac{\pi}{4}$