7.2 Simplify: \[ \frac{\sin \left(180^{\circ}-x\right) \cdot \cos \left(x-180^{\circ}\right) \cdot \tan \left(360^{\circ}-x\right)}{\sin (-x) \cdot \cos \left(450^{\circ}+x\right)} \]

\[ \textbf{Step 1: Simplify the numerator} \] We have: \[ \sin(180^\circ - x) \quad\text{and}\quad \cos(x - 180^\circ) \quad\text{and}\quad \tan(360^\circ - x). \] 1. Using the identity \(\sin(180^\circ - x) = \sin x\), we get: \[ \sin(180^\circ - x) = \sin x. \] 2. For the cosine term, using the cosine subtraction formula: \[ \cos(x - 180^\circ) = \cos x \cos 180^\circ + \sin x \sin 180^\circ. \] Knowing that \(\cos 180^\circ = -1\) and \(\sin 180^\circ = 0\), this simplifies to: \[ \cos(x - 180^\circ) = -\cos x. \] 3. For the tangent term, recall that: \[ \tan(360^\circ - x) = -\tan x. \] Now, the numerator becomes: \[ \sin x \cdot \big(-\cos x\big) \cdot \big(-\tan x\big). \] The two negatives cancel: \[ \sin x \cdot \cos x \cdot \tan x. \] \[ \textbf{Step 2: Simplify the denominator} \] The denominator is: \[ \sin(-x) \cdot \cos(450^\circ + x). \] 1. For \(\sin(-x)\), using the odd identity of sine: \[ \sin(-x) = -\sin x. \] 2. For \(\cos(450^\circ + x)\), notice: \[ 450^\circ = 360^\circ + 90^\circ. \] Thus, \[ \cos(450^\circ + x) = \cos(360^\circ + 90^\circ + x) = \cos(90^\circ + x). \] And using the identity: \[ \cos(90^\circ + x) = -\sin x, \] we have: \[ \cos(450^\circ + x) = -\sin x. \] Thus, the denominator becomes: \[ (-\sin x) \cdot (-\sin x) = \sin^2 x. \] \[ \textbf{Step 3: Combine the results} \] The overall expression now is: \[ \frac{\sin x \cdot \cos x \cdot \tan x}{\sin^2 x}. \] Substitute \(\tan x = \frac{\sin x}{\cos x}\) into the expression: \[ \frac{\sin x \cdot \cos x \cdot \frac{\sin x}{\cos x}}{\sin^2 x} = \frac{\sin^2 x}{\sin^2 x}. \] Thus, the simplified expression is: \[ 1. \]