Prove the identity tanθ+7 / tan²θ−3 = sinθcosθ+7cos²θ / 1−4cos²θ

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. [1] M1 For appropriate use of tanθ = sinθ/cosθ or tanθcosθ = sinθ at least once
2. [1] M1 For appropriate use of sin²θ + cos²θ =1 or 1+tan²θ =sec²θ
3. [1] A1 WWW Fully correct proof
4. [1] *B1 OE Using part (a) and eliminating fractions. tanθ(tanθ+7) = 5(tan²θ-3)
5. [1] DM1 Factorising or other accepted method for solving their 3-term quadratic in tanθ. Condone errors made in forming 3-term quadratic. 4tan²θ-7tanθ-15=0 => (4tanθ+5)(tanθ-3)=0
6. [1] B1 OE Independent of factorisation of the quadratic. And no extra solutions. Allow e.g. x = ... provided x = tanθ is seen. Can be implied by a correct final answer. tanθ=-1.25 and tanθ=3
7. [1] B1 AWRT No others in range 0≤θ≤180°. Ignore any answers outside this range. Maximum 3/4 if incorrect factorisation or formula. θ=71.6° and θ=128.7°