Question 927047
<pre>
tan(&#952;) = cot(30°+5&#952;)

Since tan(&#952;) = cot(90°-&#952;)

and you only want one value of &#952;,

Replace tan(&#952;) by cot(90°-&#952;)

cot(90°-&#952;) = cot(30°+5&#952;)

     90°-&#952; = 30°+5&#952;
       60° = 6&#952;
       10° = &#952;

[To get all the solutions, since tangent and cotangent
have period 180°, add n*180° to 90°-&#952; 

cot(90°-&#952;+n*180°) = cot(30°+5&#952;)

90°-&#952;+n*180° = 30°+5&#952;
  60°+n*180° = 6&#952;
   10°+n*30° = &#952; where n is any integer, positive, negative or zero. 
The solutions are given by this sequence:

..., -140°, -110°, -80°, -50°, -20°, 10°, 40°, 70°, 100°, 130°, 160°,... 

Edwin</pre>