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Given that sin²θ + cos²θ ≡ 1, show that 1 + cot²θ ≡ csc²θ - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 5

Question 7

Given that sin²θ + cos²θ ≡ 1, show that 1 + cot²θ ≡ csc²θ. (b) Solve, for 0 < θ < 180°, the equation 2 cot²θ - 9 cosecθ = 3, giving your answers to 1 decimal place.

Worked Solution & Example Answer:Given that sin²θ + cos²θ ≡ 1, show that 1 + cot²θ ≡ csc²θ - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 5

Step 1

Given that sin²θ + cos²θ ≡ 1, show that 1 + cot²θ ≡ csc²θ.

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Answer

To show that $1 + cot^2\theta \equiv csc^2\theta$ , we start with the identity given:

$sin^2\theta + cos^2\theta \equiv 1$

Dividing the entire equation by $sin^2\theta$ , we get:

$\frac{sin^2\theta}{sin^2\theta} + \frac{cos^2\theta}{sin^2\theta} \equiv \frac{1}{sin^2\theta}$

This simplifies to:

$1 + cot^2\theta \equiv csc^2\theta$

Therefore, we have shown that the given expression is correct.

Step 2

Solve, for 0 < θ < 180°, the equation 2 cot²θ - 9 cosecθ = 3, giving your answers to 1 decimal place.

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Answer

First, rewrite the equation:

$2 cot^2\theta - 9 cosec\theta - 3 = 0$

Now, express $cosec\theta$ in terms of $cot\theta$ using the identity:

$cosec^2\theta = 1 + cot^2\theta$

Then, substituting $cosec\theta = \frac{1}{sin\theta}$ and solving:

Multiply through by $sin^2\theta$ to eliminate $cosec\theta$ and create a quadratic.
Solve using the quadratic formula or factoring, finding the values of $\theta$ .

After calculation, the solutions give:

$\theta \approx 11.5° \text{ and } 168.5°$

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