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CIE 9709 2026 March Paper 12 Q5

A Level / CIE / P1

CIE 9709 2026 March Paper 12 Paper · Question 5

题目

Problem

(a) Prove the identity

11−sin⁡θ+11+sin⁡θ≡2cos⁡2θ.\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}\equiv\frac{2}{\cos^2\theta}.
(3)

(b) Hence, solve the equation

31−sin⁡θ+31+sin⁡θ=12\frac{3}{1-\sin\theta}+\frac{3}{1+\sin\theta}=12

for 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ.

(4)
题目中文翻译

(a) 证明恒等式

11−sin⁡θ+11+sin⁡θ≡2cos⁡2θ.\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}\equiv\frac{2}{\cos^2\theta}.
(3)

(b) 由此，在 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ 内解方程

31−sin⁡θ+31+sin⁡θ=12.\frac{3}{1-\sin\theta}+\frac{3}{1+\sin\theta}=12.
(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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两项通分，分子中的正弦项相消，分母用平方差公式化简。

答题过程

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Where the expressions are defined,

 11−sin⁡θ+11+sin⁡θ= (1+sin⁡θ)+(1−sin⁡θ)1−sin⁡2θ= 2cos⁡2θ.\begin{align*} &\,\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}\\ =&\,\frac{(1+\sin\theta)+(1-\sin\theta)}{1-\sin^2\theta}\\ =&\,\frac2{\cos^2\theta}. \end{align*}

(b)

解法一

思路

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利用 (a) 得到余弦的平方，再分别考虑余弦的正、负值。

答题过程

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The equation is equivalent to 6/cos⁡2θ=126/\cos^2\theta=12, so cos⁡θ=±1/2\cos\theta=\pm1/\sqrt2. The reference angle is 45∘45^\circ. Taking both signs over the stated interval gives 45∘45^\circ, 135∘135^\circ, 225∘225^\circ and 315∘315^\circ. All four have cos⁡θ≠0\cos\theta\ne0 and are in the domain of the original equation.