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(i) (sin263° + sin227°)/(cos217° + cos273°) (ii) sin 25° cos 65° + cos 25° sin 65°

(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii) sin 25° cos 65° + cos 25° sin 65°

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 8741 Points
3 months ago
(i) (sin263° + sin227°)/(cos217° + cos273°) To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes, = [sin2(90°-27°) + sin227°] / [cos2(90°-73°) + cos273°)] = (cos227° + sin227°)/(sin227° + cos273°) = 1/1 =1 (since sin2A + cos2A = 1) Therefore, (sin263° + sin227°)/(cos217° + cos273°) = 1 (ii) sin 25° cos 65° + cos 25° sin 65° To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes, = sin(90°-25°) cos 65° + cos (90°-65°) sin 65° = cos 65° cos 65° + sin 65° sin 65° = cos265° + sin265° = 1 (since sin2A + cos2A = 1) Therefore, sin 25° cos 65° + cos 25° sin 65° = 1

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