Given that 15cos²θ + 2sin²θ = 7, show that tan²θ = 8/5.

Mark scheme

Marking points from the official mark scheme (May/June 2012). The mark or mark type is shown in brackets.

1. [M1] (i) for dividing by cos² θ (15 + 2tan² θ = 7sec² θ)
2. [M1] (i) for 1/cos² θ = sec² θ (15 + 2tan² θ = 7(1 + tan² θ))
3. [M1] (i) for sec² θ = 1 + tan² θ
4. [A1] (i) for rearrangement to get required result (leading to tan² θ = 8/5)
5. OR
6. [M1] (i) for use of identity (15(1-sin²θ) + 2sin²θ = 7)
7. [M1] (i) for simplification (15 - 13sin²θ = 7 => 13sin²θ = 8)
8. [M1] (i) for use of tan θ = sin θ / cos θ (15cos²θ + 2sin²θ = 7 => 15cos²θ + 2(1-cos²θ) = 7 => 13cos²θ = 5)
9. [A1] (i) for rearrangement
10. [M1] (ii) for attempt to solve (tan θ = ±√(8/5))
11. [A1] (ii) θ = 0.902 (3 s.f.)
12. [A1] (ii) θ = 2.24 (3 s.f.)