1. **Problem Statement:** (a) Prove that $$\tan^2\theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta}$$ provided that $$1 + \cos 2\theta \neq 0$$. (b) Using the formula from part (a), find $$\tan^2\left(\frac{\pi}{8}\right)$$ in the form $$a + b\sqrt{c}$$ without a calculator. --- 2. **Formula and Important Rules:** Recall the double-angle identity for cosine: $$\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$$ Also, $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$ and $$\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$$. --- 3. **Proof of the formula (a):** Start with the right side: $$\frac{1 - \cos 2\theta}{1 + \cos 2\theta}$$ Using the identity $$\cos 2\theta = 1 - 2\sin^2\theta$$, substitute: $$\frac{1 - (1 - 2\sin^2\theta)}{1 + (1 - 2\sin^2\theta)} = \frac{1 - 1 + 2\sin^2\theta}{1 + 1 - 2\sin^2\theta} = \frac{2\sin^2\theta}{2 - 2\sin^2\theta}$$ Simplify numerator and denominator by factoring out 2: $$= \frac{\cancel{2}\sin^2\theta}{\cancel{2}(1 - \sin^2\theta)} = \frac{\sin^2\theta}{1 - \sin^2\theta}$$ Recall that $$1 - \sin^2\theta = \cos^2\theta$$, so: $$= \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$$ Thus, we have shown: $$\tan^2\theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta}$$ provided $$1 + \cos 2\theta \neq 0$$ to avoid division by zero. --- 4. **Calculate $$\tan^2\left(\frac{\pi}{8}\right)$$ using the formula (b):** Substitute $$\theta = \frac{\pi}{8}$$: $$\tan^2\left(\frac{\pi}{8}\right) = \frac{1 - \cos\left(2 \times \frac{\pi}{8}\right)}{1 + \cos\left(2 \times \frac{\pi}{8}\right)} = \frac{1 - \cos\left(\frac{\pi}{4}\right)}{1 + \cos\left(\frac{\pi}{4}\right)}$$ Recall: $$\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$ Substitute: $$= \frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}$$ Multiply numerator and denominator by 2 to clear fractions: $$= \frac{2 - \sqrt{2}}{2 + \sqrt{2}}$$ Multiply numerator and denominator by the conjugate of the denominator $$2 - \sqrt{2}$$: $$= \frac{(2 - \sqrt{2})(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{(2 - \sqrt{2})^2}{2^2 - (\sqrt{2})^2}$$ Calculate numerator: $$(2 - \sqrt{2})^2 = 2^2 - 2 \times 2 \times \sqrt{2} + (\sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2}$$ Calculate denominator: $$4 - 2 = 2$$ So: $$\tan^2\left(\frac{\pi}{8}\right) = \frac{6 - 4\sqrt{2}}{2} = 3 - 2\sqrt{2}$$ --- **Final answers:** - (a) $$\tan^2\theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta}$$ - (b) $$\tan^2\left(\frac{\pi}{8}\right) = 3 - 2\sqrt{2}$$