If \(tan^{-1}(1+x)+tan^{-1}(1-x)=\frac{\pi}{2}\) then x = ?A. 1B. -1C

1.	If \(tan^{-1}(1+x)+tan^{-1}(1-x)=\frac{\pi}{2}\) then x = ?A. 1B. -1C. 0D. \(\frac{1}{2}\)
Answer» Correct Answer is (C) 0 Since we know that tan^-1x + tan^-1 y = \(tan^{-1}(\frac{x+y}{1-xy})\) ⇒ \(tan^{-1}(1+x)+tan^{-1}(1-x)\)\(=tan^{-1}(\frac{(1+x)+(1-x)}{1-(1+x)(1-x)})\) = tan^-1 \((\frac{2}{1-(1-x^2)})\) = tan^-1\((\frac{2}{x^2})\) Here since \(tan^{-1}(1+x)+tan^{-1}(1-x)=\frac{\pi}{2}\) ⇒ tan^-1\((\frac{2}{x^2})\) = \(\frac{\pi}{2}\) ⇒ tan^-1\((\frac{2}{x^2})\)= tan^-1\((\infty)\) \((\because tan\frac{\pi}{2}=\infty)\) ⇒ \(\frac{2}{x^2}=\infty\) ⇒ \(x^2=\frac{2}{\infty}\) ⇒ x = 0

If \(tan^{-1}(1+x)+tan^{-1}(1-x)=\frac{\pi}{2}\) then x = ?A. 1B. -1C. 0D. \(\frac{1}{2}\)

Answer»

Correct Answer is (C) 0

Since we know that tan^-1x + tan^-1 y = \(tan^{-1}(\frac{x+y}{1-xy})\)

⇒ \(tan^{-1}(1+x)+tan^{-1}(1-x)\)\(=tan^{-1}(\frac{(1+x)+(1-x)}{1-(1+x)(1-x)})\)

= tan^-1 \((\frac{2}{1-(1-x^2)})\)

= tan^-1\((\frac{2}{x^2})\)

Here since \(tan^{-1}(1+x)+tan^{-1}(1-x)=\frac{\pi}{2}\)

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

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

⇒ \(\frac{2}{x^2}=\infty\)

⇒ \(x^2=\frac{2}{\infty}\)

⇒ x = 0