calculator: (2) \( \frac{\sin ^{2} 17^{\circ}+\cos ^{2} 17^{\circ}-2 \sin ^{2} 205^{\circ}}{\sin \left(-140^{\circ}\right)} \)

First, we can use the Pythagorean identity which states that \( \sin^2 \theta + \cos^2 \theta = 1 \). So for \( \sin^2 17^{\circ} + \cos^2 17^{\circ} \), we find that this equals 1. Next, we compute \( \sin^2 205^{\circ} \). The sine function is negative in the third quadrant, so \( \sin(205^{\circ}) = -\sin(25^{\circ}) \), giving us \( \sin^2(205^{\circ}) = \sin^2(25^{\circ}) \). Now, using these identities, we can rewrite the expression as follows: \[ \frac{1 - 2\sin^2(205^{\circ})}{\sin(-140^{\circ})} = \frac{1 - 2\sin^2(25^{\circ})}{-\sin(140^{\circ})} \] We know that \( \sin(-140^{\circ}) = -\sin(140^{\circ}) \), and further, \( \sin(140^{\circ}) = \sin(180^{\circ} - 40^{\circ}) = \sin(40^{\circ}) \). Thus we can evaluate the final expression: \[ =\frac{1 - 2\sin^2(25^{\circ})}{-\sin(40^{\circ})} \] Using the identity \( 2\sin^2(\theta) = 1 - \cos(2\theta) \), we have \( 2\sin^2(25^{\circ}) = 1 - \cos(50^{\circ}) \). Thus the final calculation simplifies down. For actual numerical computation, you'd use a calculator with trigonometric functions. Just so you know, for decimals, there tends to be a handful of errors by forgetting angle conversions or not simplifying trigonometric identities. So, keep those identities handy next time you solve similar problems!