Evaluating Sine

$\begin{aligned} Q & = \frac {-\sin \left(-1^\circ\right) -\sin \left(-2^\circ\right) -\sin \left(-3^\circ\right)-\cdots -\sin \left(-89^\circ\right)}{\cos \left(-1^\circ\right) +\cos \left(-2^\circ\right) + \cos \left(-3^\circ\right) + \cdots + \cos \left(-89^\circ\right)} & \small \color{#3D99F6} \text{Note that }\sin (-\theta) = - \sin \theta, \ \cos(-\theta) = \cos \theta \\ & = \frac {\sin \left(1^\circ\right) + \sin \left(2^\circ\right) + \sin \left(3^\circ\right)+\cdots +\sin \left(89^\circ\right)}{\cos \left(1^\circ\right) +\cos \left(2^\circ\right) + \cos \left(3^\circ\right) + \cdots + \cos \left(89^\circ\right)} & \small \color{#3D99F6} \text{Note that }\cos (90^\circ-\theta) = \sin \theta \\ & = \frac {\sin \left(1^\circ\right) + \sin \left(2^\circ\right) + \sin \left(3^\circ\right)+\cdots +\sin \left(89^\circ\right)}{\sin \left(89^\circ\right) + \sin \left(88^\circ\right) + \sin \left(87^\circ\right)+\cdots +\sin \left(1^\circ\right)} \\ & = \boxed{1} \end{aligned}$

This is the same as Summation of sin x from 1 to 89 all over summation of cos x from 1 to 89 degrees. Take note that by co-function, sin 1 = cos 89. Thus the numerator is the same as denominstor Therefore, the answer is 1.