Find the sum of the following :

(i) $$\frac{1}{\cos0^\circ \cos1^\circ}+\frac{1}{\cos1^\circ\cos2^\circ} +\frac{1}{\cos2^\circ \cos3^\circ}+......+\frac{1}{\cos88^\circ \cos89^\circ}$$

I tried : $$\frac{1}{\cos1^\circ}\left[\frac{\cos(1^\circ-0^\circ)}{\cos0^\circ\cos1^\circ} + \frac{\cos(2^\circ-1^\circ)}{\cos1^\circ\cos2^\circ}+...\right]$$

= $$\frac{1}{\cos1^\circ}\left[\frac{\cos1^\circ\cos0^\circ}{\cos0^\circ\cos1^\circ} - \frac{\sin1^\circ \sin0^\circ}{\sin0^\circ\cos1^\circ} + \frac{\cos2^\circ \cos1^\circ}{\cos1^\circ\cos2^\circ} -\frac{\sin2^\circ \sin1^\circ}{\cos1^\circ\cos2^\circ}...\right]$$

For this, as well, I am not getting any pattern to solve further. Please suggest, thanks.

  • 1
    $\begingroup$ Related : math.stackexchange.com/questions/425966/… $\endgroup$ Aug 10, 2013 at 3:36
  • 1
    $\begingroup$ This post is a duplicate of this: math.stackexchange.com/questions/324367/… $\endgroup$ Aug 10, 2013 at 4:30
  • $\begingroup$ @PratyushSarkar, how you have found this? I could find the related as I had solved it:) $\endgroup$ Aug 10, 2013 at 4:43
  • $\begingroup$ @labbhattacharjee I already saw that post a few months ago (look at the comment under the answer hehe). When I saw this post it looked familiar but as it was long ago I wasn't sure. So I looked through my activity in my account and found the post and by coincidence it was exactly the same. $\endgroup$ Aug 10, 2013 at 4:52

1 Answer 1



$$\frac{\sin(A-B)}{\cos A\cos B}=\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}=\tan A-\tan B$$

If $A= (n+1)^\circ,B=n^\circ$

$$\frac{\sin 1^\circ}{\cos (n+1)^\circ\cos n^\circ}=\tan(n+1)^\circ-\tan n^\circ $$

Put $n=0,1,2,\cdots,87,88$ and add to find the series to be Telescopic

  • $\begingroup$ @sultan, observe that the series in this question and the one in math.stackexchange.com/questions/425966/… are basically same $\endgroup$ Aug 10, 2013 at 4:21
  • $\begingroup$ @sultan, it won't e out of the mark to solve the following $$\sum_{1\le r\le 89}\frac1{\tan r^\circ\tan(r+1)^\circ}$$ $\endgroup$ Aug 10, 2013 at 4:46

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