Trigonometry! #126

First we have that $bc = \dfrac{\cos(x)}{k}*\dfrac{\tan(x)}{k} = \dfrac{\sin(x)}{k^{2}} = \dfrac{a}{k}.$

Next, $\dfrac{1}{ck} = \dfrac{1}{\tan(x)} = \dfrac{\cos(x)}{\sin(x)}.$

For the last term, we have $\dfrac{ak}{1 + bk} = \dfrac{\sin(x)}{1 + \cos(x)} = \dfrac{1 - \cos(x)}{\sin(x)}.$

Thus $\dfrac{1}{ck} + \dfrac{ak}{1 + bk} = \dfrac{1}{\sin(x)} = \dfrac{1}{ak},$

and so $bc + \dfrac{1}{ck} + \dfrac{ak}{1 + bk} = \dfrac{a}{k} + \dfrac{1}{ak} = \boxed{\dfrac{1}{k}\left(a + \dfrac{1}{a}\right)}.$