2022-M2-02 Posted on 11-08-2023 By app.cch No Comments on 2022-M2-02 Ans: (b) 5π12 tanθ1−cotθ+cotθ1−tanθ=tanθ1−1tanθ+1tanθ1−tanθ=tan2θtanθ−1+1tanθ(1−tanθ)=tan3θ−1tanθ(tanθ−1)=(tanθ−1)(tan2θ+tanθ+1)tanθ(tanθ−1)=tan2θ+1+tanθtanθ=sec2θ+tanθtanθ=1cos2θ÷sinθcosθ+1=1+1cosθsinθ=1+secθcscθ tanθ1−cotθ+cotθ1−tanθ=51+secθcscθ=5 , by the result of (a).1cosθsinθ=4cosθsinθ=1412sin2θ=14sin2θ=12 ∴2θ=π6 or 2θ=5π6. Hence, we have θ=π12 (rejected) or θ=5π12. Same Topic: 2022-M2-01 2023-M2-01 2023-M2-04 2023-M2-11 2022, HKDSE-M2 Tags:Trigonometry (M2)