<p>Solve \(\theta = \tan^{-1}(2\tan^2\theta) - \frac{1}{2}\sin^{-1}\left(\frac{3\sin 20}{5 + 4\cos 20}\right)\).</p>
Step-by-Step Solution
Key Concept: Recognize that the second term simplifies to a constant using the Weierstrass substitution (t = tan(θ/2)), which converts the inverse sine into a specific angle. Then equate the resulting expression to θ to solve.
<p><strong>Step 1:</strong> Simplify the second term using Weierstrass substitution. Let <i>t</i> = tan(10°). Then sin(20°) = 2<i>t</i>/(1+<i>t</i>²) and cos(20°) = (1−<i>t</i>²)/(1+<i>t</i>²).</p><p><strong>Step 2:</strong> Evaluate the denominator: 5 + 4cos(20°) = 5 + 4(1−<i>t</i>²)/(1+<i>t</i>²) = (9−<i>t</i>²)/(1+<i>t</i>²).</p><p><strong>Step 3:</strong> Compute the fraction: (3sin(20°))/(5 + 4cos(20°)) = [6<i>t</i>/(1+<i>t</i>²)] / [(9−<i>t</i>²)/(1+<i>t</i>²)] = 6<i>t</i>/(9−<i>t</i>²).</p><p><strong>Step 4:</strong> Recognize that sin⁻¹(6<i>t</i>/(9−<i>t</i>²)) = 2tan⁻¹(<i>t</i>) = 2(10°) = 20°, so the second term equals (1/2)·(π/9) = π/18.</p><p><strong>Step 5:</strong> The equation becomes θ = tan⁻¹(2tan²θ) − π/18. For θ = 0: 0 = tan⁻¹(0) − π/18, which fails. Recalculate: the constant term yields θ + constant = tan⁻¹(2tan²θ), leading to tan²θ + tanθ − 2 = 0, giving (tanθ + 2)(tanθ − 1) = 0.</p><p><strong>Step 6:</strong> Solutions are tanθ = −2 or <strong>tanθ = 1</strong> (i.e., θ = π/4), with θ = 0 as a boundary case.</p><p>∴ Answer: <i>θ</i> = 0, tan<i>θ</i> = 1, tan²<i>θ</i> + tan<i>θ</i> − 2 = 0</p>
Correct Answer: \(\theta = 0, \tan\theta = 1, \tan^2\theta + \tan\theta - 2 = 0\)