You need to use the following formula to evaluate `tan 135^o` such that:

`tan(a-b) = (tan a-tan b)/(1+ tan a*tan b)`

You may consider `135^o = 180^o- 45^o` , hence, reasoning by analogy yields:

`tan 135^o = tan(180^o - 45^o) = (tan180^o - tan 45^o)/(1 + tan180^o* tan 45^o)`

`tan 135^o = (0-1)/(1+0) = -1`

You should evaluate `cot 120^o` using the following formula such that:

`cot 120^o = 1/(tan 120^o)`

You should consider `120^o = 60^o + 60^o` , hence, using the formula `tan(2a) = (2tan a)/(1- tan^2 a)` yields:

`tan 2*60^o = (2tan60^o)/(1 - tan^2 60^o)`

`tan 2*60^o = 2sqrt3/(1 - 3) = -sqrt3`

**Hence, evaluating `tan 135^o + sin30^o*cot 120^o` yields:**

`tan 135^o + sin30^o*cot 120^o = -1 + 1/2*(-sqrt3/3)`

`tan 135^o + sin30^o*cot 120^o = -(6 + sqrt3)/6`