I really need help so please help.
write the first trigonometric function in terms of the second trigonometric function.
sec x; tan x
think of sec x as 1/cos x.
oops i take that back
big hint: sec x is the hypotenuse of a triangle with legs tan x and 1.
from
sin^2 Ø + cos^2 Ø = 1
divide each term by cos^2 Ø
sin^2 Ø/cos^2 Ø + 1 = 1/cos^2 Ø
tan^2 Ø + 1 = sec^2 Ø
sec Ø = √(tan^2 Ø + 1)
Thank you, Reiny! It helped me a lot.
To write the first trigonometric function (sec x) in terms of the second trigonometric function (tan x), we can use the definitions and relationships between trigonometric functions.
First, let's remind ourselves of the definitions:
sec x = 1 / cos x
tan x = sin x / cos x
To write sec x in terms of tan x, we need to eliminate cos x and only have sin x in the expression.
To do this, we can use the Pythagorean Identity:
sin^2 x + cos^2 x = 1
Rearranging this equation, we get:
sin^2 x = 1 - cos^2 x
Now, let's substitute this expression into the definition of tan x:
tan x = sin x / cos x
tan x = √(1 - cos^2 x) / cos x
Now, to express sec x in terms of tan x, we can substitute the expression for cos x from the definition of sec x into the equation above:
sec x = 1 / cos x
1 = cos x / sec x
cos x = 1 / sec x
Substituting this expression into the equation for tan x, we get:
tan x = √(1 - (1/sec^2 x)) / (1 / sec x)
tan x = √(1 - 1/(sec^2 x)) * sec x
Simplifying further:
tan x = √((sec^2 x - 1) / (sec^2 x)) * sec x
tan x = sec x * √(sec^2 x - 1)
Therefore, we have expressed the first trigonometric function (sec x) in terms of the second trigonometric function (tan x) as:
sec x = tan x * √(tan^2 x + 1)