Question

The value of $\tan^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right], |x|\le \frac{1}{\sqrt{2}}, x\ne0,$ is equal to [JEE(Main)-2017]

• A. $\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x^2$
• B. $\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^2$
• C. $\frac{\pi}{4}-\cos^{-1}x^2$
• D. $\frac{\pi}{4}+\cos^{-1}x^2$

Answer 1

Answer: (B)

$\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^2$

Answer 2

Let the given expression be $y$. $y = \tan^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$

The domain is given as $|x| \le \frac{1}{\sqrt{2}}$, $x \ne 0$. Squaring the inequality, we get $x^2 \le \frac{1}{2}$. Since $x \ne 0$, $x^2 \ne 0$. Also, $x^2 \ge 0$ for real $x$. Thus, $0 < x^2 \le \frac{1}{2}$.

Let's make the substitution $x^2 = \cos \theta$. Since $0 < x^2 \le \frac{1}{2}$, we have $0 < \cos \theta \le \frac{1}{2}$. As $x^2 \ge 0$, $\cos \theta \ge 0$. The range of $\theta = \cos^{-1}(x^2)$ for $x^2 \in (0, 1/2]$ is $[\cos^{-1}(1/2), \cos^{-1}(0))$, which is $[\frac{\pi}{3}, \frac{\pi}{2})$.

Now substitute $x^2 = \cos \theta$ into the expression: $y = \tan^{-1}\left[\frac{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}\right]$ Using the half-angle identities $1+\cos \theta = 2\cos^2(\theta/2)$ and $1-\cos \theta = 2\sin^2(\theta/2)$: $\sqrt{1+\cos \theta} = \sqrt{2\cos^2(\theta/2)} = \sqrt{2}|\cos(\theta/2)|$ $\sqrt{1-\cos \theta} = \sqrt{2\sin^2(\theta/2)} = \sqrt{2}|\sin(\theta/2)|$

The range of $\theta$ is $[\frac{\pi}{3}, \frac{\pi}{2})$. The range of $\theta/2$ is $[\frac{\pi}{6}, \frac{\pi}{4})$. In this range, both $\cos(\theta/2)$ and $\sin(\theta/2)$ are positive. So, $|\cos(\theta/2)| = \cos(\theta/2)$ and $|\sin(\theta/2)| = \sin(\theta/2)$.

Substituting these into the expression for $y$: $y = \tan^{-1}\left[\frac{\sqrt{2}\cos(\theta/2)+\sqrt{2}\sin(\theta/2)}{\sqrt{2}\cos(\theta/2)-\sqrt{2}\sin(\theta/2)}\right]$ $y = \tan^{-1}\left[\frac{\cos(\theta/2)+\sin(\theta/2)}{\cos(\theta/2)-\sin(\theta/2)}\right]$

To simplify the expression inside the $\tan^{-1}$, divide the numerator and the denominator by $\cos(\theta/2)$: $y = \tan^{-1}\left[\frac{\frac{\cos(\theta/2)}{\cos(\theta/2)}+\frac{\sin(\theta/2)}{\cos(\theta/2)}}{\frac{\cos(\theta/2)}{\cos(\theta/2)}-\frac{\sin(\theta/2)}{\cos(\theta/2)}}\right]$ $y = \tan^{-1}\left[\frac{1+\tan(\theta/2)}{1-\tan(\theta/2)}\right]$

This expression is in the form $\frac{1+\tan A}{1-\tan A}$, which is the expansion of $\tan(\frac{\pi}{4}+A)$. So, $y = \tan^{-1}\left[\tan(\frac{\pi}{4}+\frac{\theta}{2})\right]$.

Now we need to evaluate $\tan^{-1}(\tan(\phi))$. This is equal to $\phi$ if $\phi$ is in the principal value range of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$. The range of $\theta/2$ is $[\frac{\pi}{6}, \frac{\pi}{4})$. The range of $\frac{\pi}{4}+\frac{\theta}{2}$ is $[\frac{\pi}{4}+\frac{\pi}{6}, \frac{\pi}{4}+\frac{\pi}{4})$. $\frac{\pi}{4}+\frac{\pi}{6} = \frac{3\pi+2\pi}{12} = \frac{5\pi}{12}$. $\frac{\pi}{4}+\frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2}$. So the range of $\frac{\pi}{4}+\frac{\theta}{2}$ is $[\frac{5\pi}{12}, \frac{\pi}{2})$. This interval $[\frac{5\pi}{12}, \frac{\pi}{2})$ is a subset of the principal value range $(-\frac{\pi}{2}, \frac{\pi}{2})$. Therefore, $y = \frac{\pi}{4}+\frac{\theta}{2}$.

Substitute back $\theta$ in terms of $x^2$. We used $x^2 = \cos \theta$, so $\theta = \cos^{-1}(x^2)$. $y = \frac{\pi}{4}+\frac{\cos^{-1}(x^2)}{2}$.

Comparing this result with the given options, our result matches option (2).

The final answer is $\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^2$.