[Trigonometry] Finding the value for cosine with given the value for tangent?

When the value of tanθ is given :
sinθ/cosθ = tanθ
[√(1 - cos²θ)]/cosθ = tanθ
(1 - cos²θ)/cos²θ = tan²θ
1 - cos²θ = tan²θ cosθ
(1 + tan²θ)cos²θ = 1
cos²θ = 1/(1 + tan²θ)
cosθ = ±1/√(1 + tan²θ)
(the sign depends on the value of θ.)

Then :
cosθ = ±1/√(1 + tan²θ)
sinθ = ±√(1 - cos²θ) = ±√[1 - (1/1 + tan²θ)] = ±tanθ/√(1 + tan²θ)
cotθ = 1/tanθ
secθ = 1/cosθ = ±√(1 + tan²θ)
cscθ = 1/sinθ = ±[√(1 + tan²θ)]/tanθ
(The sign depends on the value of θ.)

Evaluate the indicated trigonometric expression using the given information (Round your answer to four decimal places). If tanθ = 1.4259 over the interval [0°, 90°], find cosθ.

When θ is in the first quadrant (over the interval [0°, 90°]), cosθ > 0
cosθ = +1/√(1 + tan²θ) = 1/√(1 + 1.4259²) = 0.5742


Evaluate the indicated trigonometric expression using the given information (Round your answer to four decimal places). If tanθ = −4.8016 over the interval [3π/2, 2π], find cscθ.

When θ is in the fourth quadrant (over the interval [3π/2, 2π]), csc < 0
csc = -[√(1 + tan²θ)]/tanθ = -[√(1 + 4.8016²)]/4.8016 = -1.0215


When the value of sinθ is given :
cosθ = ±√(1 - sin²θ)
tanθ = sinθ/cosθ = ±sinθ/√(1 - sin²θ)
cscθ = 1/sinθ
secθ = 1/cosθ = ±1/√(1 - sin²θ)
cotθ = 1/tanθ = ±[√(1 - sin²θ)]/sinθ
(the sign depends on the value of θ.)

When the value of cosθ is given :
sinθ = ±√(1 - cos²θ)
tanθ = sinθ/cosθ = ±[√(1 - cos²θ)]/cosθ
secθ = 1/cosθ
cscθ = 1/sinθ = ±1/√(1 - cos²θ)
cotθ = 1/tanθ = ±cos/[√(1 - cos²θ)]
(the sign depends on the value of θ.)

tan(θ)=1.4259 with θ acute so from a right-angled triangle
with angle θ, side o=1.4259, a=1,
h=sqrt(1^2+1.4259^2)=1.74161
so cos(θ)= a/h = 1/1.74161=0.57418~0.5742.(4DP)
This can be obtained frrom your calculator using
tan^-1(1.4250) then cos of this answer. You require
Degree Mode.

Put your calculator into Radian Mode as shown by
[3π/2, 2π]. Then θ=tan^-1(-4.8016) which is in QIV so
θ=4.91772
sinθ=-0.97899
so csc(θ)= -1/0.97899 =-1.021457 ~-1.0216 (4DP)

There are three ways to calculate this. I will show you the easiest way.

Note: when you use this arctangent function make sure your calculator is in "degree mode" or your results won't agree with me. "Radians" is angles in terms of pi, so that's the difference!

tan(Θ) = 1.4259
Calculate Θ by undoing the tangent function with the arctangent function. It's on your scientific calculator.

arctan[tan(Θ)] = arctan(1.4259)

Θ = ~ 54.95758087 degrees. # Just to check this angle is in the desired interval!
If we want to be precise for the cosine function to 4 decimal places, we will have to preserve more than 4 decimal places with the angle. Now calculate cos(Θ)

cos(Θ) is cos(54.95758087) = ~ 0.5742
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Now let's do the next problem, and we will do the same thing, but with one change.

tan(Θ) = -4.8016 ... now calculate the arctangent function:

arctan[tan(Θ)] = arctan[-4.8016]

Θ = ~ -78.23552321 degrees
Now this clearly isn't in the interval, so 360 degree rotation of the unit circle to the same point.

Θ = ~ -78.23552321 + 360 degrees = ~ 281.7644768 degrees ... this is in the interval 270 to 360 degrees which is the same interval asked for except in terms of pi, it's in degrees now.

Now we calculate csc(Θ) with the adjusted degree value.
I don't have that on my calculate so I have to do this reciprocal identity:

csc(Θ) = 1 / sin(Θ) or [sin(Θ)]^(-1) ... okay!

csc(Θ) = 1 / [sin(281.7644768)] = ~ -1.0215
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Else the other way you can solve this functions is to use the Pythagorean Trigonometry identities:

1 + [tan(Θ)]^2 = [sec(Θ)]^2
Solve the functions for sec(Θ), and you know what tan(Θ) is so substitute in that numerical value.

You never want sec(Θ) so again reciprocal identities: cos(Θ) = 1 / [sec(Θ)], as you know what sec(Θ) is numerically.

Then for the second problem you would use this identity: 1 + [cot(Θ)]^2 = [csc(Θ)]^2
Remember that cot(Θ) is 1 / tan(Θ) so ...

1 + [1 / tan(Θ)]^2 = [csc(Θ)]^2
You would probably get a positive value because you are in quadrant 1 of the unit circle where everything is positive, so in quadrant 4 as you desire cosecant would be negative. That's the only difference.

http://img.sparknotes.com/figures/0/067486b8a9659518b7099dac07405d29/quadrantsigns.gif

SOHCAHTOA
Tan is Opposite over Adjacent
Cos is Adjacent over Hypotenuse

O^2 + A^2 = H^2