Question

Q. cot⁴θ + cot²θ = 1, then sin⁴θ +sin²θ = ? 1. 1 2. 2 3. 3 4. 4 (Share solution too)

Answer 1

Answer:

ANSWER

2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)

= 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}

= 2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}

=2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)

=2-6sin²θcos²θ-3+6sin²θcos²θ

=-1

Answer 2

Answer:

2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)

= 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}

= 2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}

=2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)

=2-6sin²θcos²θ-3+6sin²θcos²θ

=-1