# Telescoping Tangents

Here's a problem I have borrowed from Imad Zak facebook group. The problem is by Mikolaj Hajduk and the solution is by Marian Dinca. The problem is simple but I am glad to add another example to the telescoping collection:

Solution

For $n\gt 1,\,$ find the value $S(n),$

$\displaystyle S(n) =\sum_{k=1}^n\frac{\sin 1}{\cos (k-1)\cos k}.$

We invoke the formula for the sine of a sum:

\displaystyle\begin{align}S(n)&=\sum_{k=1}^n\frac{\sin 1}{\cos (k-1)\cos k}\\ &=\sum_{k=1}^n\frac{\sin (k-(k-1))}{\cos (k-1)\cos k}\\ &=\sum_{k=1}^n\frac{\sin k\cos(k-1)-\sin(k-1)\cos k}{\cos (k-1)\cos k}\\ &=\sum_{k=1}^n\left(\tan k - \tan(k-1)\right)\\ &=\tan n - \tan 0\\ &=\tan n. \end{align}

### Telescoping situations

• Leibniz and Pascal Triangles
• Infinite Sums and Products
• Sum of an infinite series
• Harmonic Series And Its Parts
• A Telescoping Series
• An Inequality With an Infinite Series
• That Divergent Harmonic Series
• An Elementary Proof for Euler's Series
• $\sin 1^{\circ}+\sin {2^\circ}+\sin 3^{\circ}+\cdots+\sin 180^{\circ}=\tan 89.5^{\circ}$
• Problem 3824 from Crux Mathematicorum
• $x_n=\sin 1+\sin 3+\sin 5+\cdots+\sin (2n-1)$
• A Welcome Problem for the Year 2018
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