<p>The value of \(\sin\left[\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right) + \cos^{-1}\left(\cos\dfrac{7\pi}{3}\right)\right]\) is</p>
Step-by-Step Solution
Key Concept: Apply the range restrictions of inverse trigonometric functions: tan⁻¹ returns values in (-π/2, π/2) and cos⁻¹ returns values in [0, π]. Reduce arguments to these ranges using periodic properties.
<p><strong>Step 1: Evaluate tan⁻¹(tan(7π/6))</strong></p><p>Since 7π/6 is in the third quadrant and outside (-π/2, π/2), reduce it:</p><p>7π/6 = π + π/6, so tan(7π/6) = tan(π/6) = 1/√3</p><p>Therefore: tan⁻¹(tan(7π/6)) = tan⁻¹(1/√3) = π/6</p><p><strong>Step 2: Evaluate cos⁻¹(cos(7π/3))</strong></p><p>Reduce 7π/3 to [0, 2π]: 7π/3 = 2π + π/3</p><p>So cos(7π/3) = cos(π/3) = 1/2</p><p>Therefore: cos⁻¹(cos(7π/3)) = cos⁻¹(1/2) = π/3</p><p><strong>Step 3: Calculate the sum and apply sine</strong></p><p>sin[π/6 + π/3] = sin[π/6 + 2π/6] = sin(π/2) = 1</p><p><strong>∴ Answer: A (which equals 1)</strong></p>
Correct Answer: A