Question

How do you solve `(tanx+1)^2=sec^2x-3`?

Answer 1

`x=arctan(-3/2)+kpi`

Explanation:

Since `color(red)(tanx=sinx/cosx) and color(green)(secx=1/cosx)`, you have:

`(color(red)(sinx/cosx)+1)^2=(color(green)(1/cosx))^2-3`

`((sinx+cosx)/cosx)^2=(1-3cos^2x)/cos^2x`

`(color(blue)(sin^2x+cos^2x)+2sinxcosx)/cos^2x=(1-3cos^2x)/cos^2x`

Since `color(blue)(sin^2x+cos^2x=1)`, you have:

`cancel(color(blue)1)+2sinxcosx=cancel(color(blue)1)-3cos^2x and cosx!=0`

`2sinxcosx+3cos^2x=0 and cosx!=0`

`cosx(2sinx+3cosx)=0 and cosx!=0`

`2sinx+3cosx=0`

then you can divide by cos x to obtain:

`2color(red)(sinx/cosx)+3cosx/cosx=0`

`2color(red)tanx=-3`

`tanx=-3/2`

`x=arctan(-3/2)+kpi`

Answer 2

`-56^@31 + k180^@`

Explanation:

Use trig identity: `(tan^2 x + 1) = sec^2 x`
`(tan x + 1)^2 = tan^2 x + 1 + 2tan x = sec^2 x - 3`
`sec^2 x + 2tan x = sec^2 x - 3`
Cancel `sec^2 x` from both sides, we get
2tan x = -3
tan x = - 1.5
Calculator gives
`x = - 56^@31`.
General answers:
`x = - 56^@31 + k180^@`
Check by calculator.
x = - 56.31 --> tan x = -1. 5 --> (tan x + 1)^2 = 0.25
sec^2 x = 1/(cos^2 x) = 1/0.30 = 3.25.
sec^2 x - 3 = 3.25 - 3 = 0.25. OK