Hi there
The solve function is used to find explicit values where a given equation is satisfied. In the problem, we have the following equation:
eq1 = 1/tan(b);
This means that you would require “tan(b)” to take on an infinite value, which is not directly solvable in the typical sense. Since the solver sees the “tan(b)” function, it may realize that “tan(b)==inf” is an asymptote, and may not be able to find the solution.
A workaround is to consider a function that is not asymptotic at the expected solution. The equation can be rewritten using cos or cot functions, which are easier to deal with since they are not asymptotic at 
.
.syms b
eq1 = cot(b); % or cos(b) since cot(b)=cos(b)/sin(b)
sol = solve(eq1,'ReturnConditions', true)
Changing the definition of “1/tan(b)” to “cot(b)” resolves the problem, even though both are equivalent. This is because the solver must deal with asymptotes in the former and can avoid them in the latter. Below are plots of the tan and cot functions which illustrate asymptotic behaviour.
Hope it clarifies the problem
larush
