If tanx+secx=sqr rt 3, 0

tan x + sec x = √3

squaring on both sides

(tan x + sec x)² = √3²

tan²x+sec²x+2 tanx sec x = 3

sec²x - tan²x = 1

tan²x + 1+ tan²x+ 2tanx secx = 3

2tan²x+2tanxsecx = 3-1

tan²x+tanx secx = 2/2

tan²x+tanx secx = 1

tanx secx = 1 - tan²x

again squaring on both sides

tan²x sec²x = 1 + tan⁴x - 2 tan²x

( 1+ tan²x ) tan²x = 1 + tan⁴x - 2 tan²x

tan⁴x + tan²x = 1 + tan⁴x - 2 tan²x

3 tan²x = 1

tan x = 1/√3

x = 30.

The x value is 30.

you can check once

 

squaring both the sides,

tan²x + sec²x +2tanxsecx = 3

put sec^2x = tan²x +1

and take 2tanx common,

2tanx(tanx+sec)=2

sqrrt3(tanx)=1

therefore tanx = 1/rt3

x=30

sec²x=(√3-tanx)²

1+tan²x=3+tan²x-2√3tanx

2√3tanx=2

tanx=1/√3

x=30degree

Dear Student,
Please find below the solution to your problem.

tan x + sec x = √3
squaring on both sides
(tan x + sec x)^2 = √3^2
tan^2x + sec^2x + 2tanx sec x = 3
sec^2x - tan^2x = 1
tan^2x + 1+ tan^2x + 2tanx secx = 3
2tan^2x + 2tanxsecx = 3-1
tan^2x + tanx secx = 2/2
tan^2x + tanx secx = 1
tanx secx = 1 - tan^2x
again squaring on both sides
tan^2xsec^2x = 1 + tan^4x - 2 tan^2x
( 1+ tan^2x ) tan^2x = 1 + tan^4x - 2 tan^2x
tan^4x + tan^2x = 1 + tan^4x - 2 tan^2x
3 tan^2x = 1
tan x = 1/√3
x = 30

Thanks and Regards