Given that $\sin^2 \theta + \cos^2 \theta = 1$, show that $1 + \tan^2 \theta = \sec^2 \theta$ - Edexcel - A-Level Maths Pure - Question 3 - 2005 - Paper 5

To solve the equation, we first rewrite it using the identity for secant:

$2 \tan^2 \theta + \frac{1}{\cos \theta} = 1.$

Multiply through by $\cos \theta$:

$2 \tan^2 \theta \cos \theta + 1 = \cos \theta.$

Substituting $\tan \theta = \frac{\sin \theta}{\cos \theta}$ gives:

$2 \left(\frac{\sin^2 \theta}{\cos^2 \theta}\right) \cos \theta + 1 = \cos \theta.$

This simplifies to:

$2 \sin^2 \theta + \cos^3 \theta - \cos^2 \theta = 0.$

Next, we can use $\sin^2 \theta = 1 - \cos^2 \theta$ to rewrite this as:

$2(1 - \cos^2 \theta) + (\cos^3 \theta) - \cos^2 \theta = 0 \rightarrow 2 - 2\cos^2 \theta + \cos^3 \theta - \cos^2 \theta = 0 \rightarrow \cos^3 \theta - 3\cos^2 \theta + 2 = 0.$

Factoring gives:

$(\cos \theta - 1)(\cos^2 \theta - 2\cos \theta - 2) = 0.$

The first part gives:

$\cos \theta = 1 \implies \theta = 0^\circ.$

For the quadratic equation $\cos^2 \theta - 2\cos \theta - 2 = 0$, we can use the quadratic formula:

$\cos \theta = \frac{2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}.$

Evaluating this gives:

1. $\cos \theta = 1 - \sqrt{3} \approx 131.8^\circ,$
2. $\cos \theta = 1 + \sqrt{3} \approx 228.2^\circ.$

Thus, the final answers within the range are:

$\theta_1 \approx 0^\circ, \theta_2 \approx 131.8^\circ, \theta_3 \approx 228.2^\circ.$