Given seccant of thata = 3, use trigonometric identities to find the exact value of
a- Cos thata
b- tan squre of thata
c csc (90 degree - thata)
d- sin squre thata
secTheta=3
but secTheta= 1/cosTheta
solve for cosTheta
Then, sin^2theta+ cos^2Theta=1
solve for sinTheta.
then, Tan theta= sinTheta/cosTheta
you have it.
a) Given sec(theta) = 3, we know that sec(theta) = 1/cos(theta). So, we can solve for cos(theta):
sec(theta) = 1/cos(theta)
3 = 1/cos(theta)
To find cos(theta), we can take the reciprocal of both sides:
1/3 = cos(theta)
Therefore, cos(theta) = 1/3.
Now, we can find a - cos(theta):
a - cos(theta) = a - (1/3)
b) To find tan^2(theta), we can use the identity tan(theta) = sin(theta)/cos(theta):
tan^2(theta) = (sin(theta)/cos(theta))^2
Since we already found cos(theta) = 1/3, we can substitute it in the expression:
tan^2(theta) = (sin(theta)/(1/3))^2
tan^2(theta) = (3sin(theta))^2
So, b - tan^2(theta) = (3sin(theta))^2.
c) We are given csc(90 degrees - theta). We know that csc(theta) = 1/sin(theta). So, we can use the identity:
csc(90 degrees - theta) = 1/sin(90 degrees - theta)
Recall that sin(90 degrees - theta) = cos(theta), so we have:
csc(90 degrees - theta) = 1/cos(theta)
Using the previously found value of cos(theta) = 1/3, we have:
csc(90 degrees - theta) = 1/(1/3)
= 3
Therefore, c - csc(90 degrees - theta) = 3.
d) To find sin^2(theta), we can use the identity sin^2(theta) + cos^2(theta) = 1. Rearranging the equation, we get:
sin^2(theta) = 1 - cos^2(theta)
We can substitute the previously found value of cos(theta) = 1/3:
sin^2(theta) = 1 - (1/3)^2
= 1 - 1/9
= 8/9
So, d - sin^2(theta) = 8/9.