how to input tan theta

The quadrants are correct but the angles should be 161.565051177078° and 341.565051177078° since the angles lie in those quadrants.

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {cot}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{theta}}$$

Make sure you set your calculator to radians or degrees depending on what your problem states and solve

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = -{\mathtt{18.434\: \!948\: \!822\: \!922^{\circ}}}$$

NOTE:  inverse tan is the same as arc tan - they are different notations meaning exactly the same thing.

$$\mathrm{\ }$$

This is the answer in the 4th quadrant.

Now Tan is negative in the 2nd and 4th quadrant

2nd quad would be 180-18.43....=

$${\mathtt{180}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{198.434\: \!948\: \!822\: \!922}}$$   

WRONG

this is wrong - see my later post should be

$${\mathtt{180}}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{161.565\: \!051\: \!177\: \!078}}$$

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4th quadrant (also)

$${\mathtt{360}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{378.434\: \!948\: \!822\: \!922}}$$     

WRONG - should be

$${\mathtt{360}}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{341.565\: \!051\: \!177\: \!078}}$$

So to the nearest degree    $$\theta = 162^0\:\: or\:\:342^0 \mbox{ where }0^0\le\theta\le 360^0$$

NOTE: It may have been easier for you to just found  $$tan^{-1}(1/3)$$   and then worked it out for the required quadrants afterwards

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annoymous,

I know it is very confusing (I really dislike this notation-in my oppinion the atan notation is much nicer) but

 $$tan^{-1}(ratio)$$   DOES NOT MEAN  $$\frac{1}{tan(ratio)}$$

What do you think it should be Chris?

yes - of course - thanks Chris.

(I took away a negative number - and hence added)

this would not have happened if i had just found 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{18.434\: \!948\: \!822\: \!922^{\circ}}}$$ in th first place and worked from there.