Question

$\tan^{-1}x, x < 0$ is equal to

• A. $-\pi + \cot^{-1}\frac{1}{x}$
• B. $\sin^{-1}\frac{x}{\sqrt{1+x^2}}$
• C. $-\cos^{-1}\frac{1}{\sqrt{1+x^2}}$
• D. $\cos^{-1}\frac{1}{\sqrt{1+x^2}}$

Answer 1

Answer: (A), (B), (C)

(A), (B), (C)

Answer 2

Let $y = \tan^{-1}x$. For $x < 0$, the range of $y$ is $(-\frac{\pi}{2}, 0)$.

Option (A): For $x < 0$, $\frac{1}{x} < 0$. The principal value of $\cot^{-1}z$ for $z < 0$ lies in $(\frac{\pi}{2}, \pi)$. Thus, for $x < 0$, $\cot^{-1}\frac{1}{x} \in (\frac{\pi}{2}, \pi)$. Consequently, $-\pi + \cot^{-1}\frac{1}{x} \in (-\pi + \frac{\pi}{2}, -\pi + \pi) = (-\frac{\pi}{2}, 0)$. This range matches the range of $\tan^{-1}x$ for $x < 0$. The identity $\tan^{-1}x = -\pi + \cot^{-1}\frac{1}{x}$ holds for $x < 0$.

Option (B): We know that for $y = \tan^{-1}x$, where $-\frac{\pi}{2} < y < \frac{\pi}{2}$, $\sin y = \frac{\tan y}{\sqrt{1+\tan^2 y}} = \frac{x}{\sqrt{1+x^2}}$. Since $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $y = \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$. This identity holds for all real $x$. For $x < 0$, $\frac{x}{\sqrt{1+x^2}} \in (-1, 0)$, so $\sin^{-1}\frac{x}{\sqrt{1+x^2}} \in (-\frac{\pi}{2}, 0)$, matching the range of $\tan^{-1}x$.

Option (C): For $y = \tan^{-1}x$, $\cos y = \frac{1}{\sqrt{1+\tan^2 y}} = \frac{1}{\sqrt{1+x^2}}$ for $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Since $x < 0$, $y \in (-\frac{\pi}{2}, 0)$. Let $z = -y$. Then $z \in (0, \frac{\pi}{2})$. $\cos z = \cos(-y) = \cos y = \frac{1}{\sqrt{1+x^2}}$. Since $z \in (0, \frac{\pi}{2})$, $z = \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$. Substituting $z = -y$, we get $-y = \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$, which means $y = -\cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$. This identity holds for $x \le 0$. For $x < 0$, $\frac{1}{\sqrt{1+x^2}} \in (0, 1)$, so $\cos^{-1}\frac{1}{\sqrt{1+x^2}} \in (0, \frac{\pi}{2})$. Thus, $-\cos^{-1}\frac{1}{\sqrt{1+x^2}} \in (-\frac{\pi}{2}, 0)$, matching the range of $\tan^{-1}x$.

Option (D): For $x < 0$, $\cos^{-1}\frac{1}{\sqrt{1+x^2}} \in (0, \frac{\pi}{2})$, which does not match the range of $\tan^{-1}x$ for $x < 0$, which is $(-\frac{\pi}{2}, 0)$. Thus, option (D) is incorrect.