If $2k \sin 30^{\circ} \cos 30^{\circ} \cot 60^{\circ} = \frac{\cot^{2}30^{\circ}\sec 60^{\circ}\tan 45^{\circ}}{cosec^{2}45^{\circ}cosec^{2}30^{\circ}}$, then find the value of k.

3

$2k \sin 30^{\circ} \cos 30^{\circ} \cot 60^{\circ} = \frac{\cot^{2}30^{\circ}\sec 60^{\circ}\tan 45^{\circ}}{cosec^{2}45^{\circ}cosec^{2}30^{\circ}}$

2k × \(\frac{1}{2}\)× \(\frac{√3 }{2}\)× \(\frac{1}{√3}\)

= \(\frac{(√3)²×2 × 1  }{(√2)² × 2}\) 

2k × \(\frac{1}{4}\) = \(\frac{6  }{4}\)

k = 3