Prove that
\[2\sin \dfrac{\pi }{6}\sec \dfrac{\pi }{3}-4\sin \dfrac{5\pi }{6}\cot \dfrac{\pi }{4}=0\]

Hint: We have the trigonometric values in the given function that is \[\sin \dfrac{\pi }{6}=\dfrac{1}{2}\]and \[\sec \dfrac{\pi }{3}=2\], \[\sin \dfrac{5\pi }{6}=\dfrac{1}{2}\] because it can be written as \[\sin \left( \pi -\dfrac{\pi }{6} \right)\]so it is in second quadrant and sin is positive in that quadrant, \[\cot \dfrac{\pi }{4}=1\]

Complete step-by-step answer:
The given trigonometric function is \[2\sin \dfrac{\pi }{6}\sec \dfrac{\pi }{3}-4\sin \dfrac{5\pi }{6}\cot \dfrac{\pi }{4}\]
We know that value of trigonometric function \[\sin \dfrac{\pi }{6}\]is given by
\[\sin \dfrac{\pi }{6}=\dfrac{1}{2}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(1)
We know that the value of trigonometric function \[\sec \dfrac{\pi }{3}\]is given by
\[\sec \dfrac{\pi }{3}=2\]. . . . . . . . . . . . . . . . .. . . . . . . . . . . .. . . . . . . . . . . . .(2)
We know that the value of trigonometric function \[\sin \dfrac{5\pi }{6}\]is given by
\[\sin \dfrac{5\pi }{6}=\dfrac{1}{2}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . .(3)
We know that the value of trigonometric function \[\cot \dfrac{\pi }{4}\]is given by
\[\cot \dfrac{\pi }{4}=1\]. . . . .. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . .(4)
Substituting the above values we will get
\[2\times \dfrac{1}{2}\times 2-4\times \dfrac{1}{2}\times 1\]
\[=2-2\]
\[=0\]
Hence proved.

Note: If they have given trigonometric values with angles greater than 90 degrees then write them as \[90\pm \theta ,180\pm \theta ,270\pm \theta ,360\pm \theta \]and then find the corresponding value. Note that in the first quadrant all are positive and in the 2nd sine angles are positive and in 3rd tan are positive and in 4th cos angles are positive.