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# If tanx+secx=sqr rt 3, 0

## 4 Answers

8 years ago

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

8 years ago

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

10 months ago
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

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