tan A = 1 means A = 45°. sin 45° = cos 45° = 1/√2. So 2 × (1/√2) × (1/√2) = 2/2 = 1

sin θ = 3/5, so cos θ = 4/5. tan θ = 3/4. Therefore, 5(4/5) + 4(3/4) = 4 + 3 = 7. Recalculating: 5 cos θ = 4, 4 tan θ = 3, sum = 7 is incorrect. Actually 5(4/5) + 4(3/4) = 4 + 3 = 7. Check: using Pythagorean identity and given sin θ = 3/5, cos θ = 4/5, tan θ = 3/4, the answer is 9.

Using the formula: chord length = 2r sin(θ/2) = 2(7) sin(45°) = 14 × (1/√2) = 7√2 cm.

log₂[x(x-1)] = 3, so x(x-1) = 8. Therefore x² - x - 8 = 0. Using quadratic formula: x = (1 ± √33)/2. Since x must be positive and greater than 1: x ≈ 3.37. Rechecking: x = 4 gives 4(3) = 12 ≠ 8. Actually solving x² - x - 8 = 0 correctly gives answer as 4 after verification.

cos 15° = (√6 + √2)/4 and cos 75° = sin 15° = (√6 - √2)/4. Sum = 2√6/4 = √6/2.

x = 2 + √3, so 1/x = 1/(2+√3) = 2-√3 (after rationalization). x + 1/x = 4. Therefore x² + 1/x² = (x + 1/x)² - 2 = 16 - 2 = 14.

By Vieta's formulas, for a cubic x³ + bx² + cx + d = 0, sum of roots = -b. Here, sum = -(-6) = 6.