Question:

<p><strong>Step 1: Reduce using periodicity and even property</strong></p><p>Since f has period 8 and is even: f(-x) = f(x) and f(x+8) = f(x).</p><p>• f(-5) = f(5) = f(5-8) = f(-3) = f(3) = 2 [using f(x+8)=f(x) and f(-x)=f(x)]</p><p>• f(20) = f(20-16) = f(4) = 3 [since 20 = 2×8 + 4]</p><p>• f(-10) = f(10) = f(10-8) = f(2) = 1 [using periodicity and even property]</p><p>• f(17) = f(17-16) = f(1) = -2 [since 17 = 2×8 + 1]</p><p><strong>Step 2: Compute the inner expressions</strong></p><p>f(-5) + f(20) = 2 + 3 = 5</p><p>f(-10) + f(17) = 1 + (-2) = -1</p><p><strong>Step 3: Evaluate cos⁻¹(f(-10) + f(17))</strong></p><p>cos⁻¹(-1) = π [since cosine of π equals -1]</p><p><strong>Step 4: Evaluate the full expression</strong></p><p>tan(f(-5) + f(20)) + cos⁻¹(f(-10) + f(17)) = tan(5) + π</p><p><strong>Step 5: Apply tan⁻¹</strong></p><p>tan⁻¹(tan(5) + π) must be simplified. Since tan(5) + π ≈ 3.3796 which lies in (-π/2, π/2) ≈ (-1.571, 1.571)? No, 3.3796 > π/2.</p><p>Note: tan(5 + π) = tan(5) [periodicity of tangent], so tan⁻¹(tan(5) + π) requires recognizing that tan(5 - π) = tan(5), and 5 - π ≈ 1.8584 ∈ (-π/2, π/2).</p><p>Therefore: tan⁻¹(tan(5) + π) = tan⁻¹(tan(5 - π)) = 5 - π</p><p>∴ Answer: <strong>5 - π</strong></p>