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] , \left|x\right| < \frac{1}{2}, x \ne0, $ is equal to :

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

Answer 1

The Correct Option is A

To solve the problem of finding the value of \(\tan^{-1} \left[\frac{\sqrt{1+x^{2}} + \sqrt{1-x^{2}}}{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}\right]\) when \(\left| x \right| < \frac{1}{2}, x \neq 0\), let's proceed step by step. 

Step 1: Recognizing the Structure

The expression inside the inverse tangent function can be rewritten using the identity for the tangent of a sum. The identity is:

\(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}\)

We aim to express \(\frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\) in such a form.

Step 2: Let \(a = \sqrt{1+x^2}\) and \(b = \sqrt{1-x^2}\).

Thus, the expression becomes:

\(\frac{a + b}{a - b}\)

Step 3: Convert to a known trigonometric form

The expression \(\frac{a+b}{a-b}\) can be related to the tangent double angle identity, where

\(2 \tan u = \frac{2 \sin u \cos u}{\cos^2 u - \sin^2 u}\)

Step 4: Simplify using Double Angle Identity for Cosine

Notice that \(a = \sqrt{1 + x^2}\) and \(b = \sqrt{1 - x^2}\) imply using:

\(\cos(2u) = \cos^2 u - \sin^2 u = 1 - 2\sin^2 u = 2\cos^2 u - 1\)

Step 5: Relate back to Arccos

The input expression is perhaps in terms of \(\cos^{-1}(x^2)\) because its argument relates to cosine identity reformulated as \(\cos(2\theta) = 1 - 2x^{2}\).

Step 6: Identifying the Correct Expression in Output

Given choices express in terms of half-angle and tangent form:

\(\tan^{-1}\left(\tan(\frac{\pi}{4} + u)\right) = \frac{\pi}{4} + u\)

Thus, suitably \(u = \frac{1}{2}\cos^{-1}(x^2)\), leading us to conclude the correct form.

Final Answer: \(\frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^{2}\) satisfies the condition and is indeed the solution for the inverse tangent expression.