Q10:

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

Step 1: Substitution Strategy

To simplify the trigonometric expression within the inverse function, we introduce a substitution that utilizes the identity $\cos(2\theta) = 2\cos^2\theta - 1$. Let $x = \cos(2\theta)$.

Given the constraint $\frac{-1}{\sqrt{2}} \leq x \leq 1$, we find the range for $\theta$:

• If $x = 1$, $\cos(2\theta) = 1 \implies 2\theta = 0 \implies \theta = 0$.
• If $x = \frac{-1}{\sqrt{2}}$, $\cos(2\theta) = \frac{-1}{\sqrt{2}} \implies 2\theta = \frac{3\pi}{4} \implies \theta = \frac{3\pi}{8}$.

Thus, $0 \leq 2\theta \leq \frac{3\pi}{4}$, which implies $0 \leq \theta \leq \frac{3\pi}{8}$. [By substitution method for trigonometric simplification].

Step 2: Simplifying the Algebraic Expression

Substituting $x = \cos(2\theta)$ into the expression $\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}$:

$1 + \cos(2\theta) = 2\cos^2\theta$

$1 - \cos(2\theta) = 2\sin^2\theta$

Applying these identities [Using half-angle identities]:

$\frac{\sqrt{2\cos^2\theta} - \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta} + \sqrt{2\sin^2\theta}} = \frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta} = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$

Step 3: Applying Tangent Addition Formula

Divide the numerator and denominator by $\cos\theta$:

$\frac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right)$ [Using $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ where $\tan(\frac{\pi}{4}) = 1$].

Step 4: Evaluating the Inverse Function

We now evaluate $\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right)$.

Since $\theta \in [0, \frac{3\pi}{8}]$, then $(\frac{\pi}{4} - \theta) \in [\frac{\pi}{4} - \frac{3\pi}{8}, \frac{\pi}{4} - 0] = [-\frac{\pi}{8}, \frac{\pi}{4}]$.

This range lies within the principal value branch of $\tan^{-1}(x)$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Therefore, $\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta$.

Step 5: Back-Substitution

Since $x = \cos(2\theta)$, then $2\theta = \cos^{-1}x$, which implies $\theta = \frac{1}{2}\cos^{-1}x$.

Substituting back into the expression:

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

Final Solution: \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