Prove $\frac {1}{\cos 0^\circ \cdot \cos 1^\circ} + \ldots +\frac {1}{\cos 88^\circ \cdot \cos 89^\circ}= \frac{\cos 1^\circ}{\sin 1^\circ}$

Question

Prove the following identity:

$$\frac {1}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {1}{\cos 88^{\circ} \cdot \cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$

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After hours of trying, I wasn't able to make any significant progress, that worth mentioning. Then I finally decided to look into the solution, but it seems that author had the same problem, the listed solution is obviously wrong.

$$\sin 1^{\circ}\left(\frac {1}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {1}{\cos 88^{\circ} \cdot \cos 89^{\circ}}\right) = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$

$$\frac {\sin 1^{\circ}}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {\sin 1^{\circ}}{\cos 88^{\circ} \cdot \cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$

$$\frac {\sin (1^{\circ} - 0^{\circ})}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {\sin (89^{\circ} - 88^{\circ})}{\cos 88^{\circ} \cdot \cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$

$$\frac {\sin 1^{\circ}\cos0^{\circ} - \cos 1^{\circ}\sin 0^{\circ}}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {\sin 89^{\circ}\cos 88^{\circ} - \cos 89^{\circ}\sin 88^{\circ}}{\cos 88^{\circ} \cdot \cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$

$$\frac {\sin 1^{\circ}}{\cos 1^{\circ}} - \frac {\sin 0^{\circ}}{\cos 0^{\circ}} + \ldots +\frac {\sin 89^{\circ}}{\cos 89^{\circ}} - \frac {\sin 88^{\circ}}{\cos 88^{\circ}} + = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$

I hope I'm clear till now, beacuse everything I do was just elementary alegraic and trigonometric transformation. So now it's just telescopic series.

$$\frac {\sin 89^{\circ}}{\cos 89^{\circ}} - \frac {\sin 0^{\circ}}{\cos 0^{\circ}}= \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$

$$\frac {\sin 89^{\circ}}{\cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$

$$\frac {\cos 1^{\circ}}{\sin 1^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$

Q.E.D.

But I think this is actualy a counterproof, rather than a proof, because the identity that the author started the proof is different from the one we want. And obviously $\sin 1^{\circ} \neq 1$

My question is the identity wrong, is it a typo? Or am I missing something obvious?

Answer 1

I think it's a typo. $$ \frac{1}{\cos k \cos (k+1)}=\frac{1}{\sin 1}\frac{\sin 1}{\cos k \cos (k+1)}=\frac{1}{\sin 1} \frac{\sin(k+1)\cos(k)-\sin(k)\cos(k+1)}{\cos k \cos (k+1)} \\ =\frac{1}{\sin 1} \left(\tan(k+1)-\tan (k)\right) $$ Then $$ \sin 1\; \sum_{k=0}^{88}\frac{1}{\cos k \cos (k+1)}=\tan(89)-\tan(0)=\frac{\cos(1)}{\sin(1)} $$