Correct option is B
Given:
· A. Sin² 30° + Cos² 30° + Sin 90°
· B. Sin² 45° + Cos² 45° + Cos 90°
· C. 2Sin² 60° + 2Cos² 30° + Cos 0°
· D. 4Cos 60° + 2Cos² 30° + 5Sin² 45°
We use the standard trigonometric values:
· Sin 30° = 1/2
· Cos 30° = √3/2
· Sin 45° = 1/√2
· Cos 45° = 1/√2
· Sin 60° = √3/2
· Cos 60° = 1/2
· Sin 90° = 1
· Cos 90° = 0
· Cos 0° = 1
Calculation:
A. Sin² 30° + Cos² 30° + Sin 90°
= (1/2)² + (√3/2)² + 1
= 1/4 + 3/4 + 1
= 1 + 1
=
2
Therefore,
A → II
B. Sin² 45° + Cos² 45° + Cos 90°
= (1/√2)² + (1/√2)² + 0
= 1/2 + 1/2
=
1
Therefore,
B → III
C. 2Sin² 60° + 2Cos² 30° + Cos 0°
= 2(√3/2)² + 2(√3/2)² + 1
= 2(3/4) + 2(3/4) + 1
= 3/2 + 3/2 + 1
= 3 + 1
=
4
Therefore,
C → I
D. 4Cos 60° + 2Cos² 30° + 5Sin² 45°
= 4(1/2) + 2(√3/2)² + 5(1/√2)²
= 2 + 2(3/4) + 5(1/2)
= 2 + 3/2 + 5/2
= 2 + 4
=
6
Therefore,
D → IV
Final Answer:
A-II, B-III, C-I, D-IV
Correct option: (b)