(c) Calculate the following without the use of a calculator: (1) \( \frac{\tan 190^{\circ} \cos ^{\circ} 170^{\circ}}{\cos 290^{\circ}} \) (2) \( \frac{\sin ^{2} 17^{\circ}+\cos ^{2} 17^{\circ}-2 \sin ^{2} 205^{\circ}}{\sin \left(-140^{\circ}\right)} \) (3) \( \frac{\cos ^{2} 165^{\circ}}{\cos 60^{\circ}}-\cos ^{2} 97^{\circ}-\cos ^{2} 727^{\circ} \) (4) \( \frac{\cos 204^{\circ} \tan 315^{\circ}+2 \cos 156^{\circ} \sin ^{2} 3}{\sin ^{2}\left(-5^{\circ}\right)+\sin ^{2} 85^{\circ}-\sin ^{2} 24^{\circ}} \) (5) \( \sin 140^{\circ}-\cos 220^{\circ} \tan \left(-155^{\circ}\right) \) (6) \( \frac{\cos ^{4} 190^{\circ}-\sin ^{4}\left(-10^{\circ}\right)}{\cos 55^{\circ} \cos 325^{\circ}} \)

We wish to show that each expression can be simplified “by hand.” One acceptable answer is   (1) –[1/(2 cos 10°)],   (2) –1,   (3) (√3)/2,   (4) cos 6°⁄cos 24°,   (5) 1,   (6) 2. Below is one way to reach these answers. ────────────────────────────── Problem (1)   Evaluate [tan 190° · cos 170°]⁄[cos 290°]. Step 1. Notice that tan has period 180° so:   tan 190° = tan(190° – 180°) = tan 10°. Step 2. Write cos 170°. Since 170° = 180° – 10°, we have:   cos 170° = –cos 10°. Step 3. Express cos 290°. Note 290° = 360° – 70°, and cosine is even:   cos 290° = cos 70°. Step 4. Thus the expression becomes   [tan 10° · (–cos 10°)]⁄cos 70°     = –[tan 10° · cos 10°]⁄cos 70°. But tan 10° = sin 10°⁄cos 10°, so   –[sin 10°⁄cos 10° · cos 10°]⁄cos 70° = – sin 10°⁄cos 70°. Step 5. Since cos 70° = sin 20° (because cos 70° = sin(90°–70°) = sin 20°) we have   – sin 10°⁄ sin 20°. Now, using the double–angle formula sin 20° = 2 sin 10° cos 10°, we observe that   sin 10°⁄ sin 20° = sin 10°⁄(2 sin 10° cos 10°) = 1⁄(2 cos 10°). Thus the result is   –1⁄(2 cos 10°). That is, answer (1) = –(1/2)·sec 10°. ────────────────────────────── Problem (2)   Evaluate [sin² 17° + cos² 17° – 2 sin² 205°]⁄[sin(–140°)]. Step 1. Use the Pythagorean identity: sin²θ + cos²θ = 1. Then the numerator becomes   1 – 2 sin² 205°. Step 2. Note that 205° = 180° + 25° so sin 205° = – sin 25° and hence sin² 205° = sin² 25°. Thus the numerator is 1 – 2 sin² 25°. But 1 – 2 sin²θ equals cos 2θ. Here with θ = 25°, we get   1 – 2 sin² 25° = cos 50°. Step 3. For the denominator, sin(–140°) = – sin 140°. Since 140° = 180° – 40°, sin 140° = sin 40°. Thus the denominator is – sin 40°. Step 4. So the whole expression becomes   cos 50°⁄(– sin 40°) = – [cos 50°⁄ sin 40°]. But note that cos 50° = sin 40° (since sin(90°–50°) = sin 40°). Hence the quotient is –1. Thus answer (2) = –1. ────────────────────────────── Problem (3)   Evaluate [cos² 165°⁄cos 60°] – cos² 97° – cos² 727°. Step 1. Since cos 60° = 1/2, the first term is   cos² 165°⁄(1/2) = 2 cos² 165°. Also, 165° = 180° – 15°, so cos 165° = – cos 15° and hence cos² 165° = cos² 15°. Thus the first term is 2 cos² 15°. Step 2. For cos² 97°: 97° = 90° + 7° so cos 97° = – sin 7° and hence cos² 97° = sin² 7°. Step 3. For cos² 727°: reduce 727° by subtracting 720° (which is 2 full circles) to get 727° – 720° = 7°. So cos² 727° = cos² 7°. Step 4. The whole expression becomes   2 cos² 15° – sin² 7° – cos² 7°. But sin² 7° + cos² 7° = 1, so we have   2 cos² 15° – 1. Recognize that 2 cos²θ – 1 = cos 2θ. With θ = 15° this gives   cos 30° = (√3)/2. Thus answer (3) = (√3)/2. ────────────────────────────── Problem (4)   Evaluate [cos 204°·tan 315° + 2 cos 156°·sin² 3°]⁄[sin²(–5°) + sin² 85° – sin² 24°]. Step 1. Notice tan 315°. Since 315° = 360° – 45°, tan 315° = tan(–45°) = –1. Step 2. Write cos 204°. Since 204° = 180° + 24°, we have cos 204° = – cos