(i) sin 60° cos 30° + sin 30° cos 60°

Substituting the values:
= (√32) × (√32) + (12) × (12)
= 34 + 14 = 44 = 1.

(ii) 2 tan² 45° + cos² 30° − sin² 60°

Substituting the values:
= 2(1)² + (√32)² − (√32)²
= 2 + 34 – 34 = 2.

(iii) cos 45°sec 30° + cosec 30°

Substituting the values: cos 45° = 1√2, sec 30° = 2√3, cosec 30° = 2.
= 1√22√3 + 2 = 1√22 + 2√3√3
= 1√2 × √32(1 + √3) = √32√2(1 + √3)
To rationalize, multiply numerator and denominator by √2(√3 – 1):
= √3 (√3 – 1)2√2(√3 + 1) (√3 – 1) × √2√2 = √6(√3 – 1)2(2)(3 – 1) = 3√2 – √68.

(iv) sin 30° + tan 45° − cosec 60°sec 30° + cos 60° + cot 45°

Substituting the values:
= 12 + 1 − 2√32√3 + 12 + 1
Numerator: 32 − 2√3 = 3√3 – 42√3
Denominator: 2√3 + 32 = 4 + 3√32√3
= 3√3 – 43√3 + 4. Rationalizing:
= (3√3 – 4)(3√3 – 4)(3√3 + 4)(3√3 – 4) = (3√3 – 4)²(3√3)² – 4² = 27 – 24√3 + 1627 – 16 = 43 – 24√311.

(v) 5 cos² 60° + 4 sec² 30° − tan² 45°sin² 30° + cos² 30°

The denominator sin² 30° + cos² 30° = 1 (since sin²θ + cos²θ = 1). So we only need to evaluate the numerator:
= 5(12)² + 4(2√3)² − (1)²
= 5(14) + 4(43) − 1
= 54 + 163 − 1 = 15 + 64 – 1212 = 6712.

(i) 2 tan 30°1 + tan² 30° = ?

Evaluate: 2 (1√3)1 + (1√3)² = 2√31 + 13 = 2√343 = 2√3 × 34 = √32.
We know sin 60° = √32.
Answer: (A) sin 60°

(ii) 1 − tan² 45°1 + tan² 45° = ?

Evaluate: 1 – (1)²1 + (1)² = 1 – 11 + 1 = 02 = 0.
Answer: (D) 0

(iii) sin 2A = 2 sin A is true when A = ?

Let’s check the options:
If A = 0°: sin(2×0°) = sin(0°) = 0. And 2 sin(0°) = 2 × 0 = 0. This is true.
If A = 30°: sin(2×30°) = sin(60°) = √32. And 2 sin(30°) = 2 × 12 = 1. Not true.
Answer: (A) 0°

(iv) 2 tan 30°1 − tan² 30° = ?

Evaluate: 2 (1√3)1 − (1√3)² = 2√31 − 13 = 2√323 = 2√3 × 32 = 3√3 = √3.
We know tan 60° = √3.
Answer: (C) tan 60°

Step 1: Use the given information to form equations.
We are given tan (A + B) = √3. Since tan 60° = √3, we have:
A + B = 60° — (1)
We are given tan (A − B) = 1√3. Since tan 30° = 1√3, we have:
A − B = 30° — (2)

Step 2: Solve the system of linear equations.
Add equation (1) and equation (2):
(A + B) + (A − B) = 60° + 30°
2A = 90°
A = 45°

Step 3: Find B.
Substitute the value of A into equation (1):
45° + B = 60°
B = 60° − 45°
B = 15°

Final Answer: A = 45° and B = 15°.

(i) sin (A + B) = sin A + sin B.
False.
Justification: Let A = 30° and B = 60°. LHS = sin (30° + 60°) = sin (90°) = 1. RHS = sin 30° + sin 60° = 12 + √32 = 1 + √32. Since LHS ≠ RHS, the statement is false.

(ii) The value of sin θ increases as θ increases.
True. (for θ between 0° and 90°)
Justification: sin 0° = 0, sin 30° = 0.5, sin 45° ≈ 0.707, sin 60° ≈ 0.866, sin 90° = 1. The value is clearly increasing.

(iii) The value of cos θ increases as θ increases.
False.
Justification: cos 0° = 1, cos 30° ≈ 0.866, cos 60° = 0.5, cos 90° = 0. The value of cos θ decreases as θ increases from 0° to 90°.

(iv) sin θ = cos θ for all values of θ.
False.
Justification: This is only true when θ = 45°. For example, if θ = 30°, sin 30° = 1/2 and cos 30° = √3/2, which are not equal.

(v) cot A is not defined for A = 0°.
True.
Justification: cot A = cos Asin A. For A = 0°, cot 0° = cos 0°sin 0° = 10. Division by zero is not defined.