Given that $ ext{sin}^2 heta + ext{cos}^2 heta ext{ } ext{≡ } 1$, show that $1 + ext{cot}^2 heta ext{ } ext{≡ } ext{cosec}^2 heta$ - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 5

To solve the equation $2 \text{cot}^2\theta - 9 \text{cosec}\theta - 3 = 0$, we proceed as follows:

1. Substitute $\text{cosec}\theta = \frac{1}{\text{sin}\theta}$ and $\text{cot}\theta = \frac{\text{cos}\theta}{\text{sin}\theta}$: $2 \left(\frac{\text{cos}^2\theta}{\text{sin}^2\theta}\right) - 9 \left(\frac{1}{\text{sin}\theta}\right) = 3$

2. Rearranging gives: $2 \text{cosec}^2\theta - 9 \text{cosec}\theta - 3 = 0$

3. Let $y = \text{cosec}\theta$, then: $2y^2 - 9y - 3 = 0$

4. Solving this quadratic equation using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $y = \frac{9 \pm \sqrt{9^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}$ which simplifies to: $y = \frac{9 \pm \sqrt{81 + 24}}{4} = \frac{9 \pm \sqrt{105}}{4}$

5. Calculating the two possible values for $\text{cosec}\theta$ gives: $\text{cosec}\theta = y_1 \text{ and } y_2$

6. Converting back to $\theta$:

   • For the positive root, we find: $\theta \approx 11.5^\circ$
   • For the negative root, we find: $\theta \approx 168.5^\circ$

Thus, the solutions to the equation are approximately: $\theta = 11.5^\circ, 168.5^\circ$.