\(\rm tan^{-1}x+cot^{-1}x=\frac{\pi}{2}\) holds, when

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NDA (Held On: 18 Apr 2021) Maths Previous Year paper
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  1. x ∈ R
  2. x ∈ R - (-1, 1) only
  3. x ∈ R - {0} only
  4. x ∈ R - [-1, 1] only

Answer (Detailed Solution Below)

Option 1 : x ∈ R
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Detailed Solution

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Concept:

\(\rm tan^{-1}(x)+tan^{-1}({y})\) = \(\rm tan^{-1}\frac{x+y}{1-x\times y}\)

\(\rm cot^{-1}x=\) \(\rm tan^{-1}({1\over x})\)

Calculation:

Given, \(\rm tan^{-1}x+cot^{-1}x=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{1-x\times \frac{1}{x}} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}({\infty}) = \frac{\pi}{2}\)

This is true for all x ∈ R

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