Smplify: a) \( \frac{\sin \left(360^{\circ}-\theta\right) \cdot \cos \left(180^{\circ}+\theta\right)}{\sin 270^{\circ} \cdot \cos \left(360^{\circ}-\theta\right) \cdot \sin \left(90^{\circ}+\theta\right)} \) b) \( \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} \)

To simplify the expressions, we'll break them down step by step: a) The expression is: \[ \frac{\sin \left(360^{\circ}-\theta\right) \cdot \cos \left(180^{\circ}+\theta\right)}{\sin 270^{\circ} \cdot \cos \left(360^{\circ}-\theta\right) \cdot \sin \left(90^{\circ}+\theta\right)} \] Using trigonometric identities: - \( \sin(360^{\circ}-\theta) = -\sin(\theta) \) - \( \cos(180^{\circ}+\theta) = -\cos(\theta) \) - \( \sin(270^{\circ}) = -1 \) - \( \cos(360^{\circ}-\theta) = \cos(\theta) \) - \( \sin(90^{\circ}+\theta) = \cos(\theta) \) Now substituting these into the expression: \[ \frac{-\sin(\theta) \cdot -\cos(\theta)}{-1 \cdot \cos(\theta) \cdot \cos(\theta)} = \frac{\sin(\theta) \cos(\theta)}{-\cos^2(\theta)} \] This simplifies to: \[ -\frac{\sin(\theta)}{\cos(\theta)} = -\tan(\theta) \] b) The expression is: \[ \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} \] Using the identity \( \sin^2 x + \sin^2 (60^\circ - x) = \frac{1}{2} + \frac{1}{2} \sin(2x) \): Considering \( x = 18^\circ \) and \( 60^\circ - x = 42^\circ \), we get \( \sin^2 108^\circ = \sin^2 (90^\circ - 18^\circ) = \cos^2 18^\circ \). Thus, rewriting: \[ \cos^2 18^\circ + \sin^2 18^\circ = 1 \] Therefore, the simplified results are: a) \(-\tan(\theta)\) b) \(1\)