Answer :
Let's simplify each of these expressions step-by-step.
1. [tex]\( 12 \sin 17^{\circ} \cos 17^{\circ} \)[/tex]:
Using the double-angle identity [tex]\( \sin 2\theta = 2 \sin \theta \cos \theta \)[/tex], we have:
[tex]\[ 12 \sin 17^{\circ} \cos 17^{\circ} = 12 \left(\frac{1}{2} \sin 34^{\circ}\right) = 6 \sin 34^{\circ} \][/tex]
2. [tex]\( 2 \frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}} \)[/tex]:
Using the double-angle identity for tangent [tex]\( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \)[/tex], we have:
[tex]\[ 2 \frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}} = 2 \tan 60^{\circ} = 2 \sqrt{3} \][/tex]
3. [tex]\( 32 \cos^2 42^{\circ} - 1 \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 2 \cos^2 \theta - 1 \)[/tex], we have:
[tex]\[ 32 \cos^2 42^{\circ} - 1 = 16 (2 \cos^2 42^{\circ} - 1) = 16 \cos 84^{\circ} \][/tex]
4. [tex]\( 42 \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta \)[/tex]:
Using the double-angle identity [tex]\( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \)[/tex], we have:
[tex]\[ 42 \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta = 21 \sin \theta \][/tex]
5. [tex]\( 51 - 2 \sin^2 22.5^{\circ} \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 1 - 2 \sin^2 \theta \)[/tex], where [tex]\( 2 \times 22.5^\circ = 45^\circ \)[/tex], we have:
[tex]\[ 51 - 2 \sin^2 22.5^{\circ} = 51 - (1 - \cos 45^{\circ}) = 50 + \frac{\sqrt{2}}{2} \][/tex]
6. [tex]\( 7 \cos^2 15^{\circ} - \sin^2 15^{\circ} \)[/tex]:
Using the identity [tex]\( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \)[/tex], we have:
[tex]\[ 7 \cos^2 15^{\circ} - \sin^2 15^{\circ} = 6 \cos 30^{\circ} = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \][/tex]
7. [tex]\( 82 \sin 2A \cos 2A \)[/tex]:
Using the double-angle identity [tex]\( \sin 4\theta = 2 \sin 2\theta \cos 2\theta \)[/tex], we have:
[tex]\[ 82 \sin 2A \cos 2A = 41 \sin 4A \][/tex]
8. [tex]\( 6 \frac{2 \tan \frac{1}{2} \theta}{1 - \tan^2 \frac{1}{2} \theta} \)[/tex]:
Using the double-angle identity [tex]\( \tan \theta = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}} \)[/tex], we have:
[tex]\[ 6 \frac{2 \tan \frac{1}{2} \theta}{1 - \tan^2 \frac{1}{2} \theta} = 6 \tan \theta \][/tex]
9. [tex]\( 101 - 2 \sin^2 3 \theta \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 1 - 2 \sin^2 \theta \)[/tex], we have:
[tex]\[ 101 - 2 \sin^2 3 \theta = 101 - (1 - \cos 6\theta) = 100 + \cos 6\theta \][/tex]
10. [tex]\( 11 \frac{\tan 2\theta}{1 - \tan^2 2\theta} \)[/tex]:
Using the identity [tex]\( \tan 4\theta = \frac{\tan 2\theta}{1 - \tan^2 2\theta} \)[/tex], we have:
[tex]\[ 11 \frac{\tan 2\theta}{1 - \tan^2 2\theta} = 11 \tan 4\theta \][/tex]
11. [tex]\( 2 \cos^2 \frac{1}{2} \theta - 1 \)[/tex]:
Using the identity [tex]\( \cos \theta = 2 \cos^2 \frac{\theta}{2} - 1 \)[/tex], we have:
[tex]\[ 2 \cos^2 \frac{1}{2} \theta - 1 = \cos \theta \][/tex]
12. [tex]\( 13 \frac{1 - \tan^2 20^{\circ}}{\tan 20^{\circ}} \)[/tex]:
Using the identity [tex]\( \cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan 20^\circ} \)[/tex], we have:
[tex]\[ 13 \frac{1 - \tan^2 20^{\circ}}{\tan 20^{\circ}} = 13 \cot 40^{\circ} \][/tex]
13. [tex]\( \sec \theta \csc \theta \)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \][/tex]
[tex]\[ \sec \theta \csc \theta = \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta} = \frac{1}{\sin \theta \cos \theta} \][/tex]
14. [tex]\( 12 \sin x \cos x \)[/tex]:
Using the double-angle identity [tex]\( \sin 2\theta = 2 \sin \theta \cos \theta \)[/tex], we have:
[tex]\[ 12 \sin x \cos x = 6 \sin 2x \][/tex]
15. [tex]\( 151 - 2 \sin^2 \frac{1}{2} \theta \)[/tex]:
Using the identity [tex]\( \cos \theta = 1 - 2 \sin^2 \frac{\theta}{2} \)[/tex], we have:
[tex]\[ 151 - 2 \sin^2 \frac{1}{2} \theta = 151 - (1 - \cos \theta) = 150 + \cos \theta \][/tex]
1. [tex]\( 12 \sin 17^{\circ} \cos 17^{\circ} \)[/tex]:
Using the double-angle identity [tex]\( \sin 2\theta = 2 \sin \theta \cos \theta \)[/tex], we have:
[tex]\[ 12 \sin 17^{\circ} \cos 17^{\circ} = 12 \left(\frac{1}{2} \sin 34^{\circ}\right) = 6 \sin 34^{\circ} \][/tex]
2. [tex]\( 2 \frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}} \)[/tex]:
Using the double-angle identity for tangent [tex]\( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \)[/tex], we have:
[tex]\[ 2 \frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}} = 2 \tan 60^{\circ} = 2 \sqrt{3} \][/tex]
3. [tex]\( 32 \cos^2 42^{\circ} - 1 \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 2 \cos^2 \theta - 1 \)[/tex], we have:
[tex]\[ 32 \cos^2 42^{\circ} - 1 = 16 (2 \cos^2 42^{\circ} - 1) = 16 \cos 84^{\circ} \][/tex]
4. [tex]\( 42 \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta \)[/tex]:
Using the double-angle identity [tex]\( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \)[/tex], we have:
[tex]\[ 42 \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta = 21 \sin \theta \][/tex]
5. [tex]\( 51 - 2 \sin^2 22.5^{\circ} \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 1 - 2 \sin^2 \theta \)[/tex], where [tex]\( 2 \times 22.5^\circ = 45^\circ \)[/tex], we have:
[tex]\[ 51 - 2 \sin^2 22.5^{\circ} = 51 - (1 - \cos 45^{\circ}) = 50 + \frac{\sqrt{2}}{2} \][/tex]
6. [tex]\( 7 \cos^2 15^{\circ} - \sin^2 15^{\circ} \)[/tex]:
Using the identity [tex]\( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \)[/tex], we have:
[tex]\[ 7 \cos^2 15^{\circ} - \sin^2 15^{\circ} = 6 \cos 30^{\circ} = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \][/tex]
7. [tex]\( 82 \sin 2A \cos 2A \)[/tex]:
Using the double-angle identity [tex]\( \sin 4\theta = 2 \sin 2\theta \cos 2\theta \)[/tex], we have:
[tex]\[ 82 \sin 2A \cos 2A = 41 \sin 4A \][/tex]
8. [tex]\( 6 \frac{2 \tan \frac{1}{2} \theta}{1 - \tan^2 \frac{1}{2} \theta} \)[/tex]:
Using the double-angle identity [tex]\( \tan \theta = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}} \)[/tex], we have:
[tex]\[ 6 \frac{2 \tan \frac{1}{2} \theta}{1 - \tan^2 \frac{1}{2} \theta} = 6 \tan \theta \][/tex]
9. [tex]\( 101 - 2 \sin^2 3 \theta \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 1 - 2 \sin^2 \theta \)[/tex], we have:
[tex]\[ 101 - 2 \sin^2 3 \theta = 101 - (1 - \cos 6\theta) = 100 + \cos 6\theta \][/tex]
10. [tex]\( 11 \frac{\tan 2\theta}{1 - \tan^2 2\theta} \)[/tex]:
Using the identity [tex]\( \tan 4\theta = \frac{\tan 2\theta}{1 - \tan^2 2\theta} \)[/tex], we have:
[tex]\[ 11 \frac{\tan 2\theta}{1 - \tan^2 2\theta} = 11 \tan 4\theta \][/tex]
11. [tex]\( 2 \cos^2 \frac{1}{2} \theta - 1 \)[/tex]:
Using the identity [tex]\( \cos \theta = 2 \cos^2 \frac{\theta}{2} - 1 \)[/tex], we have:
[tex]\[ 2 \cos^2 \frac{1}{2} \theta - 1 = \cos \theta \][/tex]
12. [tex]\( 13 \frac{1 - \tan^2 20^{\circ}}{\tan 20^{\circ}} \)[/tex]:
Using the identity [tex]\( \cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan 20^\circ} \)[/tex], we have:
[tex]\[ 13 \frac{1 - \tan^2 20^{\circ}}{\tan 20^{\circ}} = 13 \cot 40^{\circ} \][/tex]
13. [tex]\( \sec \theta \csc \theta \)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \][/tex]
[tex]\[ \sec \theta \csc \theta = \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta} = \frac{1}{\sin \theta \cos \theta} \][/tex]
14. [tex]\( 12 \sin x \cos x \)[/tex]:
Using the double-angle identity [tex]\( \sin 2\theta = 2 \sin \theta \cos \theta \)[/tex], we have:
[tex]\[ 12 \sin x \cos x = 6 \sin 2x \][/tex]
15. [tex]\( 151 - 2 \sin^2 \frac{1}{2} \theta \)[/tex]:
Using the identity [tex]\( \cos \theta = 1 - 2 \sin^2 \frac{\theta}{2} \)[/tex], we have:
[tex]\[ 151 - 2 \sin^2 \frac{1}{2} \theta = 151 - (1 - \cos \theta) = 150 + \cos \theta \][/tex]
