Respuesta :
use the identity
sec^2x = 1 + tan^2 x
- so sec x = sqrt(1 + tan^2 x) then:-
tan x + sqrt( 1 + tan^2 x) = 1
sqrt ( 1 + tan^2 x) = 1 - tan x
1 + tan^2 x = 1 + tan^2x - 2 tan x
0 = -2 tanx
tan x = 0
x = 0, π
π is an extraneous root because sec 180 = -1
So the answer is 0 degrees
sec^2x = 1 + tan^2 x
- so sec x = sqrt(1 + tan^2 x) then:-
tan x + sqrt( 1 + tan^2 x) = 1
sqrt ( 1 + tan^2 x) = 1 - tan x
1 + tan^2 x = 1 + tan^2x - 2 tan x
0 = -2 tanx
tan x = 0
x = 0, π
π is an extraneous root because sec 180 = -1
So the answer is 0 degrees
All the solutions in the interval [0, 2π) for tan x + sec x = 1 is 0.
What are trigonometric ratios?
"Trigonometric ratios are the ratios of the sides of a right angle triangle with respect to any acute angle of the triangle."
Identities used
sec²x -tan²x = 1
According to the question,
Given,
tan x + sec x =1
⇒sec x = 1 - tan x _____(1)
Using trigonometric identity we have,
sec²x = 1 + tan²x
⇒sec x = √ 1+ tan²x _______(2)
Compare (1) and (2) we get,
√ 1+ tan²x = 1 - tan x
Squaring both the sides of trigonometric equation we get,
1+ tan²x = ( 1 - tan x)²
⇒ 1+ tan²x = 1+ tan²x -2tanx
⇒ -2tan x = 0
⇒ tan x = 0
⇒ x = 0, π
But sec π = -1 which is not possible.
So π is extraneous root.
Therefore, x = 0.
Hence, solutions in the interval [0, 2π) for tan x + sec x = 1 is x= 0.
Learn more about trigonometric identities here
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