1. **Problem Statement:** Given the equation $$\log(\sec A - \tan A) + \log(\sec A + \tan A) - \log(\sin^2 A + \cos^2 A) = \log k,$$ find the value of $k$. 2. **Formula and Important Rules:** - Recall the logarithm property: $$\log a + \log b = \log (ab).$$ - Also, $$\log a - \log b = \log \left(\frac{a}{b}\right).$$ - The Pythagorean identity: $$\sin^2 A + \cos^2 A = 1.$$ 3. **Step-by-step Solution:** - Combine the first two logarithms: $$\log(\sec A - \tan A) + \log(\sec A + \tan A) = \log\left((\sec A - \tan A)(\sec A + \tan A)\right).$$ - Using the difference of squares formula: $$(\sec A - \tan A)(\sec A + \tan A) = \sec^2 A - \tan^2 A.$$ - Substitute back into the equation: $$\log\left(\sec^2 A - \tan^2 A\right) - \log(\sin^2 A + \cos^2 A) = \log k.$$ - Using the Pythagorean identity for sine and cosine: $$\sin^2 A + \cos^2 A = 1,$$ so the equation becomes: $$\log\left(\sec^2 A - \tan^2 A\right) - \log 1 = \log k.$$ - Since $$\log 1 = 0,$$ this simplifies to: $$\log\left(\sec^2 A - \tan^2 A\right) = \log k.$$ - Recall the identity: $$\sec^2 A - \tan^2 A = 1.$$ - Therefore: $$\log 1 = \log k \implies k = 1.$$ 4. **Final Answer:** $$\boxed{1}$$