Correct Answer - Option 1 : x ∈ R
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
