[tex]\ \ \dfrac{2}{\sqrt{3} \cos (x) + \sin(x)} = \sec\left(\frac{\pi}{6} - x\right) [/tex]
Right-hand side
[tex]\text{RHS} = \sec\left(\dfrac{\pi}{6} - x\right)[/tex]
Since [tex]\sec x = \frac{1}{\cos x}[/tex], it follows that
[tex]\sec\left(\frac{\pi}{6} - x\right) = \dfrac{1}{\cos\left(\frac{\pi}{6} - x\right) }[/tex]
So we can rewrite
[tex] \begin{aligned} \text{RHS} &= \sec\left(\dfrac{\pi}{6} - x\right) \\ &= \dfrac{1}{\cos\left(\frac{\pi}{6} - x\right)} \end{aligned} [/tex]
We have a cosine difference identity for the denominator:
[tex]\begin{aligned} \cos(A-B) &= \cos A \cos B + \sin A \sin B \\ \cos\left(\tfrac{\pi}{6} - x\right) &= \cos\left(\tfrac{\pi}{6}\right) \cos (x) + \sin\left(\tfrac{\pi}{6}\right)\sin(x) \end{aligned}[/tex]
Since [tex]\sin\left(\tfrac{\pi}{6}\right) = 1/2[/tex] and [tex]\cos\left(\tfrac{\pi}{6}\right) = \sqrt{3}/2[/tex], we have
[tex]\begin{aligned}\cos\left(\tfrac{\pi}{6} - x\right) &= \cos\left(\tfrac{\pi}{6}\right) \cos (x) + \sin\left(\tfrac{\pi}{6}\right)\sin(x) \\ &= \tfrac{\sqrt{3}}{2}\cos (x) + \tfrac{1}{2}\sin(x) \end{aligned}[/tex]
Using this in the right-hand side
[tex]\begin{aligned} \text{RHS} &= \dfrac{1}{\cos\left(\frac{\pi}{6} + x\right)} \\ &= \frac{1}{\tfrac{\sqrt{3}}{2}\cos (x) + \tfrac{1}{2}\sin(x)} \end{aligned}[/tex]
Notice how we have tiny denominators of 2.If we multiply the numerator and denominator of the entire fraction, we will deal with those twos, as 2 will distribute and cancel.
[tex]\begin{aligned} \text{RHS} &= \frac{1}{\tfrac{\sqrt{3}}{2}\cos (x) + \tfrac{1}{2}\sin(x)} \\ &=\frac{2 \cdot(1)}{2 \cdot \left(\tfrac{\sqrt{3}}{2}\cos (x) + \tfrac{1}{2}\sin(x)\right)} \\ &= \frac{2}{\sqrt{3} \cos (x) + \sin(x)} \\ &= \text{LHS} \end{aligned}[/tex]
