Question

Prove that \(\tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x\), where \(-\frac{1}{\sqrt{2}} \leq x \leq 1\).

Hint:

Recognizing the pattern for trigonometric substitution is key. For expressions involving \(\sqrt{a^2-x^2}\), use \(x=a\sin\theta\) or \(x=a\cos\theta\). For \(\sqrt{a^2+x^2}\), use \(x=a\tan\theta\). For expressions like \(\sqrt{1\pm x}\), the substitution \(x=\cos(2\theta)\) is extremely useful.

Answer 1

Step 1: Understanding the Concept:
This problem requires proving an identity involving inverse trigonometric functions. The key is to simplify the complex argument of the \(\tan^{-1}\) function using a suitable trigonometric substitution.
Step 2: Key Formula or Approach:
The presence of terms \(\sqrt{1+x}\) and \(\sqrt{1-x}\) suggests the substitution \(x = \cos(2\theta)\). This is because: \[ 1 + \cos(2\theta) = 2\cos^2\theta \] \[ 1 - \cos(2\theta) = 2\sin^2\theta \] We will also use the identity \(\tan(\frac{\pi}{4} - \theta) = \frac{1-\tan\theta}{1+\tan\theta}\).
Step 3: Detailed Explanation:
Let the Left Hand Side (LHS) be \(y\).
\[ y = \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) \] Let \(x = \cos(2\theta)\). This implies \(2\theta = \cos^{-1}x\), so \(\theta = \frac{1}{2}\cos^{-1}x\).
Now, let's determine the range of \(\theta\). Given \(-\frac{1}{\sqrt{2}} \leq x \leq 1\), we have \(-\frac{1}{\sqrt{2}} \leq \cos(2\theta) \leq 1\).
The range of \(\cos^{-1}\) is \([0, \pi]\). Therefore, \(\cos^{-1}(1) \leq 2\theta \leq \cos^{-1}(-\frac{1}{\sqrt{2}})\). \[ 0 \leq 2\theta \leq \frac{3\pi}{4} \implies 0 \leq \theta \leq \frac{3\pi}{8} \] In this interval, both \(\sin\theta\) and \(\cos\theta\) are non-negative.
Now substitute \(x=\cos(2\theta)\) into the expression: \[ \sqrt{1+x} = \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2\theta} = \sqrt{2}\cos\theta \] \[ \sqrt{1-x} = \sqrt{1-\cos(2\theta)} = \sqrt{2\sin^2\theta} = \sqrt{2}\sin\theta \] Substituting these into the LHS: \[ y = \tan^{-1}\left(\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right) = \tan^{-1}\left(\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}\right) \] Divide the numerator and denominator inside the parenthesis by \(\cos\theta\) (which is non-zero in the given range).
\[ y = \tan^{-1}\left(\frac{1 - \tan\theta}{1 + \tan\theta}\right) \] Using the identity \(\tan(\frac{\pi}{4})=1\), we can write: \[ y = \tan^{-1}\left(\frac{\tan(\frac{\pi}{4}) - \tan\theta}{1 + \tan(\frac{\pi}{4})\tan\theta}\right) \] This is the formula for \(\tan(A-B)\). \[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) \] Since \(0 \leq \theta \leq \frac{3\pi}{8}\), we have \(-\frac{\pi}{8} \leq \frac{\pi}{4}-\theta \leq \frac{\pi}{4}\). This range is within the principal value branch of \(\tan^{-1}\), i.e., \((-\frac{\pi}{2}, \frac{\pi}{2})\).
So, we can simplify \(y = \frac{\pi}{4} - \theta\).
Substitute back \(\theta = \frac{1}{2}\cos^{-1}x\).
\[ y = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x \] Thus, LHS = RHS. Step 4: Final Answer:
The given identity is proved.