5.) \( \frac{\cos \left(90^{\circ}-2 x\right)}{2-2 \sin (-x) \cdot \cos \left(90^{\circ}+x\right)} \)

To simplify the expression: \[ \frac{\cos \left(90^{\circ}-2 x\right)}{2-2 \sin (-x) \cdot \cos \left(90^{\circ}+x\right)} \] let's break it down step by step. ### Step 1: Simplify the Numerator \[ \cos \left(90^{\circ} - 2x\right) \] Using the co-function identity: \[ \cos(90^{\circ} - \theta) = \sin \theta \] So, \[ \cos \left(90^{\circ} - 2x\right) = \sin(2x) \] And using the double-angle identity: \[ \sin(2x) = 2 \sin x \cos x \] ### Step 2: Simplify the Denominator \[ 2 - 2 \sin (-x) \cdot \cos \left(90^{\circ} + x\right) \] First, recall that: \[ \sin(-x) = -\sin x \] \[ \cos(90^{\circ} + x) = -\sin x \] Substituting these: \[ 2 - 2 (-\sin x) \cdot (-\sin x) = 2 - 2 \sin^2 x \] Factor out the 2: \[ 2(1 - \sin^2 x) \] Using the Pythagorean identity: \[ 1 - \sin^2 x = \cos^2 x \] So, \[ 2 \cos^2 x \] ### Step 3: Combine Numerator and Denominator Now, substitute the simplified numerator and denominator back into the original expression: \[ \frac{2 \sin x \cos x}{2 \cos^2 x} \] Cancel out the common factors: \[ \frac{\sin x}{\cos x} = \tan x \] ### Final Simplified Form \[ \tan x \] **Answer:** After simplifying, the expression is equal to tangent x. \(\tan x\)