Find $x$ if $\frac1{\sin1°\sin2°}+\frac1{\sin2°\sin3°}+\cdots+\frac1{\sin89°\sin90°} = \cot x\cdot\csc x$

Question

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?

Answer 1

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 $x=1^{\circ}.$

Answer 2

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 $(0°,90°)$.