Question

$\tan^{-1}\frac{1}{\sqrt{x^2-1}}=$

• A. $\frac{\pi}{2} + \operatorname{cosec}^{-1}x$
• B. $\frac{\pi}{2} + \sec^{-1}x$
• C. $\operatorname{cosec}^{-1}x$
• D. $\sec^{-1}x$

Answer 1

Answer: (C)

$\operatorname{cosec}^{-1}x$

Answer 2

Let the expression be $y = \tan^{-1}\frac{1}{\sqrt{x^2-1}}$. The domain of the expression is $|x| > 1$.

Case 1: $x > 1$. Substitute $x = \sec\theta$, where $\theta \in (0, \pi/2)$. $y = \tan^{-1}\frac{1}{\sqrt{\sec^2\theta-1}} = \tan^{-1}\frac{1}{\sqrt{\tan^2\theta}} = \tan^{-1}\frac{1}{\tan\theta} = \tan^{-1}(\cot\theta) = \tan^{-1}(\tan(\pi/2 - \theta))$. Since $\theta \in (0, \pi/2)$, $\pi/2 - \theta \in (0, \pi/2)$, which is in the range of $\tan^{-1}$. So, $y = \pi/2 - \theta$. Since $x = \sec\theta$, $\theta = \sec^{-1}x$. $y = \pi/2 - \sec^{-1}x$. For $x > 1$, $\operatorname{cosec}^{-1}x + \sec^{-1}x = \pi/2$, so $\pi/2 - \sec^{-1}x = \operatorname{cosec}^{-1}x$. Thus, for $x > 1$, $y = \operatorname{cosec}^{-1}x$.

Case 2: $x < -1$. Substitute $x = \sec\theta$, where $\theta \in (\pi/2, \pi)$. $y = \tan^{-1}\frac{1}{\sqrt{\sec^2\theta-1}} = \tan^{-1}\frac{1}{\sqrt{\tan^2\theta}} = \tan^{-1}\frac{1}{-\tan\theta} = \tan^{-1}(-\cot\theta) = \tan^{-1}(\tan(\theta - \pi/2))$. Since $\theta \in (\pi/2, \pi)$, $\theta - \pi/2 \in (0, \pi/2)$, which is in the range of $\tan^{-1}$. So, $y = \theta - \pi/2$. Since $x = \sec\theta$, $\theta = \sec^{-1}x$. $y = \sec^{-1}x - \pi/2$. For $x < -1$, $\operatorname{cosec}^{-1}x + \sec^{-1}x = \pi/2$, so $\pi/2 - \sec^{-1}x = \operatorname{cosec}^{-1}x$, or $\sec^{-1}x - \pi/2 = -\operatorname{cosec}^{-1}x$. Thus, for $x < -1$, $y = -\operatorname{cosec}^{-1}x$.

Since option (c) $\operatorname{cosec}^{-1}x$ matches the result for $x > 1$, and is one of the given options, we choose it, assuming the question implicitly refers to the case $x > 1$ or accepts partial matching.