if sin theta =1/2 and cos theta= √(3)/2. find tha values of tan θ, cot θ, sec θ and cscθ.
can someone show me the answer to tanθ and cotθ
my answer to secθ is 2/√3 and cscθ is 2
Both sin theta and cos theta are positive, so theta must be in quad I
construct your right-angled triangle with opposite as 1 and hypotenuse as 2
and use Pythagoras to find the adjacent to be √3 .
So now simply use the definitions of the trig ratios to find the other 4 missing ratios.
eg sec theta = hypotenuse/adjacent = 2/√3
btw, memorize SOH CAH TOA
let me know what you get
my answers are
secθ is 2/√3
cscθ is 2
i really don't know how to get tanθ and cot θ
the tanθ is 22.20
cot θ is 1/22.20
How can you possibly come up with those numbers ?? They are not even sides of the triangle.
the others are correct
tan θ = opp/adj = 1/√3 and cot θ = √3
i did tan^-1 ((1/2)/(√(3)/2)) on my calculator
tan^-1 ((1/2)/(√(3)/2)) would have given you the angle
Of course that angle would obviously be 30 degrees, not 22.2
sin A = 1/2 = Y/r.
Cos A = sqrt3/2 = X/r.
x = sqrt 3, Y = 1, r = 2.
Tan A = Y/X = 1/sqrt3.
Cot A = 1/Tan A = sqrt3/1 = sqrt 3.
sec A = 1/Cos A = 2/sqrt 3.
csc A = 1/sin A = 2/1 = 2.
To find the values of tan θ, cot θ, sec θ, and csc θ given sin θ = 1/2 and cos θ = √(3)/2, we can use the fundamental trigonometric identities.
1. Tangent (tan θ) is defined as the ratio of the sine to the cosine of an angle:
tan θ = sin θ / cos θ
Substituting the values of sin θ and cos θ, we get:
tan θ = (1/2) / (√3/2)
= 1 / √3
= (√3 / 3)
2. Cotangent (cot θ) is the reciprocal of the tangent:
cot θ = 1 / tan θ
Substituting the value of tan θ, we get:
cot θ = 1 / (√3 / 3)
= 3 / √3
= (√3)
3. Secant (sec θ) is the reciprocal of the cosine:
sec θ = 1 / cos θ
Substituting the value of cos θ, we get:
sec θ = 1 / (√3/2)
= 2 / √3
= (2√3 / 3)
4. Cosecant (csc θ) is the reciprocal of the sine:
csc θ = 1 / sin θ
Substituting the value of sin θ, we get:
csc θ = 1 / (1/2)
= 2
Therefore, the values of tan θ, cot θ, sec θ, and csc θ are (√3 / 3), (√3), (2√3 / 3), and 2 respectively.