Given that cotθ = -12/5

We can use Pythagorean theorem AC² = AB² + BC²

AC² = (12)² + (5)²

AC² = 144 + 25

AC² = 169

AC = 13

Now cosθ = BC/AC= 12/13 (cotθ is 2nd,3rd quadrature negative)

5, 12, 13 is a Pythagorean triple, i.e. 5^2 + 12^2 = 13^2

so

cotθ = -12/5 = (-12/13) / (5/13) = cosθ/sinθ, giving cosθ=-12/13, use this if θ is in QII

or

cotθ = 12/(-5) = (12/13) / (-5/13) = cosθ/sinθ, giving cosθ = 12/13, use this if θ is in QIV

cot θ = (cos θ) / (sin θ) .... definition of cotangent (in circle-based trig) or basic identity

cot² θ = (cos² θ) / (sin² θ) .... square both sides

= (cos² θ) / (1 - cos² θ) .... Pythagorean identity for sin² θ

(cot² θ)(1 - cos² θ) + cos² θ .... "solve" for cos²

cot² θ = cos² θ + (cot² θ)(cos² θ) = (1 + cot²θ)(cos² θ)

cos² θ = (cot² θ)/(1 + cot² θ)

That works for any angle θ. Now you can plug in the given value of cot θ:

cos² θ = (12²/5²) / (1 + 12²/5²)

= 12² / (5² + 12²) .... then multiply by 5²/5² to clear nested fractions

= 12² / 13² .... remembering that 12² + 5² = 13² is a Pythagorean triple

So cos θ must be ±12/13, after taking square roots. You need extra information (like what quadrant) to decide which.

tan Ө = - 5 / 12

Ө in quadrant 2 or quadrant 4

cos Ө = - 12 / 13 _______quadrant 2

cos Ө = 12 / 13 ________quadrant 4

(-12)² + 5² = 169

cosθ = ±12/√169 = ±12/13

Cotangent = adjacent/opposite, so 12 is the length of the adjacent side to whatever angle is in play and 5 is the length of the side opposite that angle. By the Pythagorean Theorem:

5^2 + 12^2 = x^2

25 + 144 = x^2

169 = x^2

+/- 13 = x (normally, the minus is disregarded).

The hypotenuse is 13, and since cosine = adjacent/hypotenuse...

cos(theta) = +/- 12/13.

***Note that whatever sign cos(theta) has, sin(theta) has the opposite sign. Both answers are possible because you aren't given the quadrant where theta is.***

Cot (Th) = Cos(Th / Sin(Th) = -12/5 = adjacent / opposite. = (a/h) / (o/h)

Using Pythagoras we an find the hypotenuse (h)

h^2 = (-12)^2 + 5^2

h^2 = 144 + 25 = 169

h^2 = 16

h = 13

Hence

Cos(Th) = -12/13