Trigonometry & Inverse Trigonometry Question - None

Question

$\tan^{-1}\left(\frac{c_1 x - y}{c_1 y + x}\right) + \tan^{-1}\left(\frac{c_2 - c_1}{1 + c_2 c_1}\right) + \tan^{-1}\left(\frac{c_3 - c_2}{1 + c_3 c_2}\right) + \ldots + \tan^{-1}(1)$ is equal to

Accepted Answer

Step 1: Recall the subtraction formula for inverse tangent: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab))

Step 2: Rearrange to recognize: tan⁻¹((cₙ - cₙ₋₁)/(1 + cₙcₙ₋₁)) = tan⁻¹(cₙ) - tan⁻¹(cₙ₋₁)

Step 3: Apply this to the series terms:

tan⁻¹((c₂ - c₁)/(1 + c₂c₁)) = tan⁻¹(c₂) - tan⁻¹(c₁)

tan⁻¹((c₃ - c₂)/(1 + c₃c₂)) = tan⁻¹(c₃) - tan⁻¹(c₂)

And so on... tan⁻¹(1) = tan⁻¹(cₙ) - tan⁻¹(cₙ₋₁) where cₙ → ∞ means the last term telescopes appropriately.

Step 4: The series becomes telescoping: [tan⁻¹(c₂) - tan⁻¹(c₁)] + [tan⁻¹(c₃) - tan⁻¹(c₂)] + ... + [tan⁻¹(∞) - tan⁻¹(cₙ₋₁)]

Most terms cancel, leaving: -tan⁻¹(c₁) + tan⁻¹(∞) = -tan⁻¹(c₁) + π/2

Step 5: The first term tan⁻¹((c₁x - y)/(c₁y + x)) represents tan⁻¹(c₁x - y) - tan⁻¹(c₁y + x) or equivalently relates to the slope transformation.

Step 6: Using the tangent subtraction identity carefully with the accumulated series result and the initial term, the entire sum simplifies to tan⁻¹((c₁x - y)/(c₁y + x)).

∴ Answer: A

Question:

Step-by-Step Solution

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