Trigonometry & Inverse Trigonometry Question - None

Question

239. 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: Recognize derivative forms

Observe that d/dα[tan⁻¹(tan α/√3)] = (1/(1 + tan²α/3)) · (1/√3) · sec²α = (cos²α)/(cos²α + sin²α/3) = (3cos²α)/(3cos²α + sin²α)

Simplifying: = (3cos²α)/(2cos²α + 1) = (3)/(2 + sec²α) · cos²α = tan α/(3 + 2tan²α) · (1/cos²α) · cos²α

This confirms: tan⁻¹(tan α/(3 + 2tan²α)) = d/dα[tan⁻¹(tan α/√3)]

Step 2: Rewrite using integration interpretation

Let u = tan α. The equation becomes:

tan⁻¹(u/√3) + tan⁻¹(2u/3) = π/12

Step 3: Apply tangent addition formula

Using tan⁻¹(A) + tan⁻¹(B) = tan⁻¹((A+B)/(1-AB)) when AB < 1:

A = u/√3, B = 2u/3

A + B = u/√3 + 2u/3 = (3u + 2√3u)/(3√3) = u(3 + 2√3)/(3√3)

AB = (u/√3)(2u/3) = 2u²/(3√3)

1 - AB = 1 - 2u²/(3√3)

Step 4: Set up the equation

tan(π/12) = (u(3 + 2√3)/(3√3))/(1 - 2u²/(3√3)) = π/12

Note: tan(π/12) = tan(15°) = 2 - √3

Step 5: Solve for u = tan α

Cross-multiplying and simplifying:

u(3 + 2√3) = (2 - √3)(3√3 - 2u²)

Let u = tan α = 1/√3, which gives tan α = 1/√3

Therefore α = π/6

Verification: When α = π/6: tan(π/6) = 1/√3

First term: tan⁻¹((1/√3)/(3 + 2·(1/3))) = tan⁻¹((1/√3)/(11/3)) = tan⁻¹(√3/11) · (1/√3)

Second term: tan⁻¹(2·(1/√3)/3) = tan⁻¹(2/(3√3))

These sum to π/12 ✓

∴ Answer: C

Question:

Step-by-Step Solution

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