# Computing $\frac{\sin(3)}{\cos(0)\cos(3)}+\frac{\sin(3)}{\cos(3)\cos(6)} +\cdots+\frac{\sin(3)}{\cos(42)\cos(45)}$

I'm trying to compute the following sum

$$\frac{\sin(3)}{\cos(0)\cos(3)}+\frac{\sin(3)}{\cos(3)\cos(6)} +\cdots+\frac{\sin(3)}{\cos(42)\cos(45)}$$

$$\cos(3)$$ and $$\sin(3)$$ do seem to be common term in all. I'm not sure how to pull them out unless we have a multiplication in denominator $$\cos(A)\cos(B) = \frac{1}{2}\left (\cos(A-B)+\cos(A+B)\right )$$

Hint: Write $$\sin3^{\circ}=\sin(3^{\circ}-0^{\circ})=\sin3^{\circ}\cos0^{\circ}-\cos3^{\circ}\sin0^{\circ}$$ $$\\$$
$$\sin3^{\circ}=\sin(6^{\circ}-3^{\circ})=\sin6^{\circ}\cos3^{\circ}-\cos6^{\circ}\sin3^{\circ}$$ $$\\$$
... $$\\$$
$$\sin3^{\circ}=\sin(45^{\circ}-42^{\circ})=\sin45^{\circ}\cos42^{\circ}-\cos45^{\circ}\sin42^{\circ}$$
$$\sum_{k=0}^{14}\frac{\sin3^{\circ}}{\cos3^{\circ}k\cos3^{\circ}(k+1)}=\sum_{k=0}^{14}(\tan3^{\circ}(k+1)-\tan3^{\circ}k)=\tan45^{\circ}-\tan0^{\circ}=1.$$