Question

prove this question please​

Answer 1

Step-by-step explanation:

Given: (1 + cot²Ø) + (1 + 1/cot²Ø) = 1/(sin²Ø - sin⁴Ø). We need to prove that LHS is equal to the RHS.

Taking LHS,

→ (1 + cot²Ø) + (1 + 1/cot²Ø)

Now, cotØ = cosØ/sinØ

→ (1 + cos²Ø/sin²Ø) + (1 + 1/(cos²Ø/sin²Ø))

→ (sin²Ø + cos²Ø)/sin²Ø + (1 + 1 × sin²Ø/cos²Ø)

Since, sin²Ø + cos²Ø = 1. So,

→ 1/sin²Ø + (1 + sin²Ø/cos²Ø)

→ 1/sin²Ø + (cos²Ø + sin²Ø)/cos²Ø

→ 1/sin²Ø + 1/cos²Ø

→ (cos²Ø + sin²Ø)/sin²Ø.cos²Ø

→ 1/sin²Ø.cos²Ø

Used formula: cos²Ø = 1 - sin²Ø

→ 1/(1 - sin²Ø).sin²Ø

→ 1/(sin²Ø - sin⁴Ø)

Hence, proved.

TRIGONOMETRY FORMULAS:

secØ = 1/cosØ, cosØ = 1/secØ

cosecØ = 1/sinØ, sinØ = 1/cosecØ

tanØ = sinØ/cosØ, cotØ = cosØ/sinØ

cotØ = 1/tanØ, tanØ = 1/cotØ

sin²Ø + cos²Ø = 1

tan²Ø + 1 = sec²Ø

1 + cot²Ø = cosec²Ø

sinØ = cos(90° - Ø)

cosØ = sin(90° - Ø)

tanØ = cot(90° - Ø)