1. **State the problem:** Simplify the expression $$R = \frac{(\sin \alpha + \cos \alpha)^2 + (\sin \alpha - \cos \alpha)^2}{(\sec \alpha + \cos \alpha)^2 - (\cos \alpha - \sec \alpha)^2}$$ 2. **Recall formulas and identities:** - Square of sums and differences: $ (a+b)^2 = a^2 + 2ab + b^2 $, $ (a-b)^2 = a^2 - 2ab + b^2 $ - Pythagorean identity: $\sin^2 \alpha + \cos^2 \alpha = 1$ - Definition: $\sec \alpha = \frac{1}{\cos \alpha}$ 3. **Simplify numerator:** $$ (\sin \alpha + \cos \alpha)^2 + (\sin \alpha - \cos \alpha)^2 $$ Expanding each: $$ (\sin^2 \alpha + 2 \sin \alpha \cos \alpha + \cos^2 \alpha) + (\sin^2 \alpha - 2 \sin \alpha \cos \alpha + \cos^2 \alpha) $$ Combine like terms: $$ (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \alpha + \cos^2 \alpha) + (2 \sin \alpha \cos \alpha - 2 \sin \alpha \cos \alpha) $$ The $2 \sin \alpha \cos \alpha$ terms cancel out: $$ 1 + 1 = 2 $$ 4. **Simplify denominator:** $$ (\sec \alpha + \cos \alpha)^2 - (\cos \alpha - \sec \alpha)^2 $$ Use difference of squares formula: $a^2 - b^2 = (a-b)(a+b)$ Let $a = \sec \alpha + \cos \alpha$, $b = \cos \alpha - \sec \alpha$ Then denominator = $$ ( (\sec \alpha + \cos \alpha) - (\cos \alpha - \sec \alpha) ) \times ( (\sec \alpha + \cos \alpha) + (\cos \alpha - \sec \alpha) ) $$ Simplify each factor: First factor: $$ \sec \alpha + \cos \alpha - \cos \alpha + \sec \alpha = 2 \sec \alpha $$ Second factor: $$ \sec \alpha + \cos \alpha + \cos \alpha - \sec \alpha = 2 \cos \alpha $$ So denominator = $$ 2 \sec \alpha \times 2 \cos \alpha = 4 \sec \alpha \cos \alpha $$ Since $\sec \alpha = \frac{1}{\cos \alpha}$, then $$ 4 \sec \alpha \cos \alpha = 4 \times \frac{1}{\cos \alpha} \times \cos \alpha = 4 \cancel{\times \frac{1}{\cos \alpha} \times \cos \alpha} $$ The $\cos \alpha$ cancels with $\frac{1}{\cos \alpha}$: $$ = 4 $$ 5. **Combine numerator and denominator:** $$ R = \frac{2}{4} = \frac{\cancel{2}}{2 \times \cancel{2}} = \frac{1}{2} $$ **Final answer:** $$ R = \frac{1}{2} $$