Chứng minh đẳng thức sau

To prove the given trigonometric identity:

\[ 8 \sin x = \frac{\sqrt{3}}{\cos x} + \frac{1}{\sin x} \]

Let's start by simplifying the right-hand side (RHS) of the equation.

First, combine the terms on the RHS over a common denominator:

\[ \frac{\sqrt{3}}{\cos x} + \frac{1}{\sin x} = \frac{\sqrt{3} \sin x + \cos x}{\cos x \sin x} \]

Now, let's rewrite the left-hand side (LHS) in terms of \(\sin x\) and \(\cos x\):

\[ 8 \sin x \]

We need to show that:

\[ 8 \sin x = \frac{\sqrt{3} \sin x + \cos x}{\cos x \sin x} \]

Multiply both sides by \(\cos x \sin x\) to clear the denominator on the RHS:

\[ 8 \sin x \cdot \cos x \sin x = \sqrt{3} \sin x + \cos x \]

Simplify the LHS:

\[ 8 \sin^2 x \cos x = \sqrt{3} \sin x + \cos x \]

Now, let's check if there are any specific values of \(x\) that satisfy this equation. For simplicity, let's try \(x = \frac{\pi}{6}\):

For \(x = \frac{\pi}{6}\):

\[ \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \]
\[ \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

Substitute these values into the equation:

\[ 8 \left(\frac{1}{2}\right)^2 \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \]

Simplify:

\[ 8 \left(\frac{1}{4}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \]
\[ 8 \left(\frac{\sqrt{3}}{8}\right) = \sqrt{3} \]
\[ \sqrt{3} = \sqrt{3} \]

Since both sides are equal, the equation holds true for \(x = \frac{\pi}{6}\).

However, to prove the identity for all \(x\), we would need to show that the equation holds for any \(x\). The above steps show that the equation holds for a specific value of \(x\), but a general proof would require more comprehensive steps, potentially involving trigonometric identities or other algebraic manipulations.