Prove the identity sin θ/(sin θ + cos θ) + cos θ/(sin θ - cos θ) = (tan²θ + 1)/(tan²θ - 1).

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

1. M1 Sight of a correct common denominator, either in one or two fractions, condone missing brackets if recovered. In the numerator condone ± sign errors only. (sinθ(sinθ-cosθ)+cosθ(sinθ+cosθ)) / ((sinθ+cosθ)(sinθ-cosθ))
2. DM1 Divide throughout by cos²θ. (sin²θ+cos²θ)/(sin²θ-cos²θ)
3. A1 tan²θ + 1 / tan²θ - 1 AG
4. M1 Replace tan² with sin²θ/cos²θ and multiply top and bottom by cos²θ. Condone ± sign errors. (sin²θ/cos²θ + 1) / (sin²θ/cos²θ - 1)
5. DM1 Simplify as far as sinθ=, or cosθ=. 1 = 2sin²θ - 2(1-sin²θ)
6. A1 A1 for either correct answer then A1FT For their second value being π – (their first) and no others in range 0<θ<π , both values must be exact and in radians. θ = π/3, 2π/3
7. A1 FT SC: B1 for θ=60°,120° or 0.333π,0.667π AWRT. or 1.05, 2.09 AWRT.
8. M1 Equate expression from (a) to 2 and clear fraction. (tan²θ + 1)/(tan²θ - 1) = 2 => tan²θ + 1 = 2(tan²θ - 1)
9. DM1 Simplify as far as tanθ= . May be implied by a correct final answer in degrees or radians. tan²θ = 3 => tanθ = [±]√3