Is given that \( p=\cos \alpha+\sin \alpha \) and \( q=\cos \alpha-\sin \alpha \) 1 Determine the following trigonometric ratios in terms of \( p \) and/or \( q \) (a) \( \cos 2 \alpha \) (b) \( \tan \alpha \) (c) \( \frac{1+\sin 2 a}{\cos 2 a} \)

To solve the problem, we start with the given expressions for \( p \) and \( q \): \[ p = \cos \alpha + \sin \alpha \] \[ q = \cos \alpha - \sin \alpha \] We will derive the required trigonometric ratios step by step. ### (a) Determine \( \cos 2\alpha \) Using the double angle formula for cosine, we have: \[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \] We can express \( \cos^2 \alpha \) and \( \sin^2 \alpha \) in terms of \( p \) and \( q \): 1. From \( p \) and \( q \), we can find \( \cos^2 \alpha \) and \( \sin^2 \alpha \): - Adding \( p \) and \( q \): \[ p + q = (\cos \alpha + \sin \alpha) + (\cos \alpha - \sin \alpha) = 2\cos \alpha \implies \cos \alpha = \frac{p + q}{2} \] - Subtracting \( q \) from \( p \): \[ p - q = (\cos \alpha + \sin \alpha) - (\cos \alpha - \sin \alpha) = 2\sin \alpha \implies \sin \alpha = \frac{p - q}{2} \] 2. Now, we can find \( \cos^2 \alpha \) and \( \sin^2 \alpha \): \[ \cos^2 \alpha = \left(\frac{p + q}{2}\right)^2 = \frac{(p + q)^2}{4} \] \[ \sin^2 \alpha = \left(\frac{p - q}{2}\right)^2 = \frac{(p - q)^2}{4} \] 3. Substitute these into the double angle formula: \[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = \frac{(p + q)^2}{4} - \frac{(p - q)^2}{4} \] 4. Simplifying this: \[ \cos 2\alpha = \frac{(p + q)^2 - (p - q)^2}{4} \] \[ = \frac{(p^2 + 2pq + q^2) - (p^2 - 2pq + q^2)}{4} \] \[ = \frac{4pq}{4} = pq \] Thus, we have: \[ \cos 2\alpha = pq \] ### (b) Determine \( \tan \alpha \) Using the definitions of sine and cosine, we have: \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{p - q}{2}}{\frac{p + q}{2}} = \frac{p - q}{p + q} \] Thus, we have: \[ \tan \alpha = \frac{p - q}{p + q} \] ### (c) Determine \( \frac{1 + \sin 2\alpha}{\cos 2\alpha} \) Using the double angle formulas, we know: \[ \sin 2\alpha = 2\sin \alpha \cos \alpha \] Substituting the expressions for \( \sin \alpha \) and \( \cos \alpha \): \[ \sin 2\alpha = 2 \left(\frac{p - q}{2}\right) \left(\frac{p + q}{2}\right) = \frac{(p - q)(p + q)}{2} \] Now, substituting into the expression: \[ \frac{1 + \sin 2\alpha}{\cos 2\alpha} = \frac{1 + \frac{(p - q)(p + q)}{2}}{pq} \] Simplifying this: \[ = \frac{2 + (p - q)(p + q)}{2pq} \] \[ = \frac{2 + p^2 - q^2}{2pq} \] Thus, we have: \[ \frac{1 + \sin 2\alpha}{\cos 2\alpha} = \frac{2 + p^2 - q^2}{2pq} \] ### Summary of Results 1. \( \cos 2\alpha = pq \) 2. \( \tan \alpha = \frac{p - q}{p + q} \) 3. \( \frac{1 + \sin 2\alpha}{\cos 2\alpha} = \frac{2 + p^2 - q^2}{2pq} \)