\( \frac { \cos x } { \operatorname { cosec } + \sin x } - \frac { \cos x } { \cos x \sin x } = \operatorname { tas } 2 x \)

Let’s start by simplifying the given equation step by step. We have: \[ \frac{\cos x}{\csc x + \sin x} - \frac{\cos x}{\cos x \sin x} = \tan 2x \] First, recall that \(\csc x = \frac{1}{\sin x}\). So, we can rewrite the first term: \[ \frac{\cos x}{\frac{1}{\sin x} + \sin x} = \frac{\cos x}{\frac{1 + \sin^2 x}{\sin x}} = \frac{\cos x \sin x}{1 + \sin^2 x} \] Now, substituting this back into the equation gives us: \[ \frac{\cos x \sin x}{1 + \sin^2 x} - \frac{\cos x}{\cos x \sin x} = \tan 2x \] The second term simplifies to: \[ \frac{\cos x}{\cos x \sin x} = \frac{1}{\sin x} = \csc x \] Now, the equation becomes: \[ \frac{\cos x \sin x}{1 + \sin^2 x} - \csc x = \tan 2x \] Converting it back, we can express: \[ \csc x = \frac{1}{\sin x} \] Combining the fractions on the left, we can find a common denominator: \[ \frac{\cos x \sin^2 x - 1 - \sin^2 x}{(1 + \sin^2 x) \sin x} \] The final expression now can be compared to \(\tan 2x\), which is \(\frac{2 \tan x}{1 - \tan^2 x}\). By equating both sides, you can simplify further and analyze for specific x values to find solutions or necessary conditions for \(x\).