Trigonometry & Inverse Trigonometry Question - None

Question

Let $a \in \left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$ such that $\tan^{-1}\!\left(\dfrac{\tan\alpha}{3 + 2\tan^2\alpha}\right) + \tan^{-1}\!\left(\dfrac{2\tan\alpha}{3}\right) = \dfrac{\pi}{12}$, then $\alpha$ equals:

Accepted Answer

Step 1: Let t = tan(α). We need to apply tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1-xy)).

Here x = t/(3+2t²) and y = 2t/3.

Step 2: Calculate x + y:

x + y = t/(3+2t²) + 2t/3 = [3t + 2t(3+2t²)]/[3(3+2t²)] = [3t + 6t + 4t³]/[3(3+2t²)] = [9t + 4t³]/[3(3+2t²)] = t(9+4t²)/[3(3+2t²)]

Step 3: Calculate 1 - xy:

xy = [t/(3+2t²)] · [2t/3] = 2t²/[3(3+2t²)]

1 - xy = [3(3+2t²) - 2t²]/[3(3+2t²)] = [9 + 6t² - 2t²]/[3(3+2t²)] = [9 + 4t²]/[3(3+2t²)]

Step 4: Apply the addition formula:

tan⁻¹((x+y)/(1-xy)) = tan⁻¹({t(9+4t²)/[3(3+2t²)]}/{[9+4t²]/[3(3+2t²)]}) = tan⁻¹(t) = tan⁻¹(tan α) = α

Step 5: Therefore: α = π/12

∴ Answer: C

Question:

Step-by-Step Solution

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