Question

sin²+sin⁴=1 then cot²+cot⁴=?

Answer 1

Given:

→ sin²∅ + sin⁴∅ = 1

→ sin²∅(1 + sin²∅)

→ (1 - cos²∅)(1 + 1 - cos²∅) = 1

→ (1 - cos²∅)(2 - cos²∅) = 1

→2 - cos²∅ - 2cos²∅ + cos⁴∅ = 1

→ 1 - cos²∅ + 1 - 2cos²∅ + cos⁴∅ = 1

→ sin²∅ + (1 - cos²∅) = 1

Case 1:

→ sin²∅ + sin²∅ = 1

→ sin²∅ = ½

Case 2:

→ (1 - cos²∅) + (1 - cos²∅) = 1

→ 2 - 2 cos²∅ = 1

→ - 2 cos²∅ = - 1

→ cos²∅ = ½

→ cot²∅ + cot⁴∅ = (cos∅/sin∅)² + (cos∅/sin∅)⁴

→ cos²∅/sin²∅ + (cos²∅/sin²∅)²

→ (1/2)/(1/2) + [(1/2)/(1/2)]²

→ 1 + 1² = 2

Identities Used

• (1 - cos²∅) = sin²∅
• cot∅ = cos∅/sin∅

Answer: 2

Answer 2

Step-by-step explanation:

sin^2A + sin^4A = 1

→sin^4A = cos^2A

→cot^2A + cot^4A = cot^2A•cosec^2A = cos^2A•cosec^2A/sin^2A

=cos^2A/sin^4A

=1