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This question already has an answer here:

If $$\dfrac1{\sin1°\sin2°}+\dfrac1{\sin2°\sin3°}+\cdots+\dfrac1{\sin89°\sin90°} = \cot x\cdot\csc x$$ and $x\in(0°,90°)$, find $x$.

I tried writing in $\sec$ form but nothing clicked. Any ideas?

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marked as duplicate by lab bhattacharjee trigonometry Jun 27 '18 at 14:35

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  • $\begingroup$ Not sure if this is a useful observation, but we do have $$ \frac{1}{\sin n^\circ \sin(n+1)^\circ} = \frac{1}{\sin (n+1)^\circ - \sin n^\circ}\left[\frac{1}{\sin n^\circ} - \frac{1}{\sin(n+1)^\circ}\right] \\ = \frac{1}{2 \cos(n + \frac 12)^\circ \cdot \sin (\frac 12)^\circ} \left[\frac{1}{\sin n^\circ} - \frac{1}{\sin(n+1)^\circ}\right] $$ $\endgroup$ – Omnomnomnom Jun 27 '18 at 5:11
  • $\begingroup$ This divided the right angle into 90 pieces. Try dividing into only 2 or 3 pieces and find the corresponding answer there $\endgroup$ – Empy2 Jun 27 '18 at 6:06
  • $\begingroup$ See also math.stackexchange.com/questions/464031/… $\endgroup$ – lab bhattacharjee Jun 27 '18 at 14:36
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It's $$\frac{1}{\sin1^{\circ}}(\cot1^{\circ}-\cot2^{\circ}+\cot2^{\circ}-\cot3^{\circ}+...+\cot89^{\circ}-\cot90^{\circ})=\frac{\cos{x}}{\sin^2{x}}$$ or $$\frac{\cos{x}}{\sin^2{x}}=\frac{\cos{1^{\circ}}}{\sin^2{1^{\circ}}}$$ in since $f(x)=\frac{\cos{x}}{\sin^2{x}}$ decreases on $\left(0,\frac{\pi}{2}\right),$ we obtain $x=1^{\circ}.$

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Note that $$\dfrac{\sin{(1°)}}{\sin(k°)\sin((k+1)°)}=\dfrac{\sin((k+1)°-k°)}{\sin(k°)\sin((k+1)°)}=\cot(k°)-\cot((k+1)°).$$ Hence $$\sum_{k=1}^{89}\frac1{\sin(k°)\sin((k+1)°)}= \frac{\cot(1°)-\overbrace{\cot(90°)}^{=0}}{\sin(1°)}=\cot (1°)\cdot\csc (1°).$$ Now show that $x=1°$ is the only solution of the given equation in the interval $(0°,90°)$.

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  • $\begingroup$ Well done ! I delete my junk. Cheers. $\endgroup$ – Claude Leibovici Jun 27 '18 at 6:23

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