$\text{Let}$ $\alpha =\frac{1}{\sin 60°\sin 61°}+\frac{1}{\sin 62°\sin 63°}+\dots \dots +\frac{1}{\sin 118°\sin 119°}$

$\text{Then the value of}$  ${\left(\frac{\mathrm{cosec}1°}{\alpha }\right)}^2$  $\text{is}\text{\_\_\_\_\_\_\_\_}$

The correct answer is 3.

Let's analyze the given expression for $\alpha$:
$\alpha =\frac{1}{\sin 60°\sin 61°}+\frac{1}{\sin 62°\sin 63°}+\cdots +\frac{1}{\sin 118°\sin 119°}$

Notice that this sum is not a standard telescoping series because the terms are skipped: $(60°,61°)$, then $(62°,63°)$, and so on, up to $(118°,119°)$.
Let us write the general term as:
$T_k=\frac{1}{\sin \left(2k\right)°\sin (2k+1)°}$
for $k=30,31,\cdots ,59$.

To simplify each term, we can multiply and divide by $\sin 1°$:
$T_k=\frac{1}{\sin 1°}·\frac{\sin 1°}{\sin \left(2k\right)°\sin (2k+1)°}$
Since $1°=(2k+1)°-\left(2k\right)°$, we have:
$T_k=\frac{1}{\sin 1°}·\frac{\sin \left(\right(2k+1)°-(2k)°)}{\sin \left(2k\right)°\sin (2k+1)°}$

Using the identity $\sin (A-B)=\sin A\cos B-\cos A\sin B$:
$T_k=\frac{1}{\sin 1°}·\left(\cot \left(2k\right)°-\cot (2k+1)°\right)$

Thus, we can write $\alpha$ as:
$\alpha =\mathrm{cosec}1°\sum _{k=30}^{59}\left(\cot \left(2k\right)°-\cot (2k+1)°\right)$

This sum expands to:
$S=(\cot 60°-\cot 61°)+(\cot 62°-\cot 63°)+\cdots +(\cot 118°-\cot 119°)$

Since $\cot (180°-\theta )=-\cot \theta$, we can group the terms from both ends.
Specifically, let us pair the terms:
- The last negative term is $-\cot 119°=-(-\cot 61°)=\cot 61°$ which cancels out with the first negative term $-\cot 61°$.
- The last positive term is $\cot 118°=-\cot 62°$ which cancels out with the second positive term $\cot 62°$.
In general, for each term in the sum:
$\cot \left(2k\right)°-\cot (2k+1)°$
Let $2k\text{'}=178-2k$. The corresponding paired term from the end is:
$\cot (178-2k)°-\cot (179-2k)°=-\cot (2k+2)°+\cot (2k+1)°$

When we add these up, all terms of the form $\cot \theta$ for $\theta =61°,62°,\cdots ,119°$ cancel in pairs because:
$-\cot (2k+1)°$ cancels with $-\cot (179-2k)°=\cot (2k+1)°$.
$\cot \left(2k\right)°$ cancels with $\cot (180-2k)°=-\cot \left(2k\right)°$ for all $2k\ne 60°$ (since $180-60°=120°$, which is not in the set of indices, the term $\cot 60°$ remains unpaired and does not cancel).

Thus, the only term that does not cancel is the very first term:
$S=\cot 60°$

We know that:
$\cot 60°=\frac{1}{\sqrt{3}}$

Substituting this back into the expression for $\alpha$:
$\alpha =\mathrm{cosec}1°·\frac{1}{\sqrt{3}}$

We want to find the value of:
${\left(\frac{\mathrm{cosec}1°}{\alpha }\right)}^2={\left(\frac{\mathrm{cosec}1°}{\mathrm{cosec}1°·{\displaystyle \frac{1}{\sqrt{3}}}}\right)}^2={\left(\sqrt{3}\right)}^2=3$

Thus, the final value is 3.