Q.$\tan^{-1} \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x$, $-\frac{1}{\sqrt{2}} \le x \le 1$ [Hint: Put $x = \cos 2\theta$] Solve the following equations:

Concept: Inverse Tangent Identity — The expression inside $\tan^{-1}$ simplifies to $\tan(\frac{\pi}{4} - \theta)$ when $x = \cos 2\theta$.

Step 1: Substitute $x = \cos 2\theta$, so $\sqrt{1+x} = \sqrt{1+\cos 2\theta} = \sqrt{2\cos^2\theta} = \sqrt{2}\,|\cos\theta|$.

For the given domain $-\frac{1}{\sqrt{2}} \le x \le 1$, we have $0 \le 2\theta \le \frac{3\pi}{4}$, so $\theta \in [0, \frac{3\pi}{8}]$, where $\cos\theta \ge 0$. Hence $|\cos\theta| = \cos\theta$.

Step 2: Similarly, $\sqrt{1-x} = \sqrt{1-\cos 2\theta} = \sqrt{2\sin^2\theta} = \sqrt{2}\,\sin\theta$ (since $\theta \ge 0$, $\sin\theta \ge 0$).

Step 3: The fraction becomes

Step 4: Therefore,

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

We need to show that for $-\frac{1}{\sqrt{2}} \le x \le 1$,

The hint suggests putting $x = \cos 2\theta$. Why? Because expressions like $\sqrt{1 \pm x}$ become $\sqrt{1 \pm \cos 2\theta}$, which simplify beautifully using half-angle formulas. This is the classic trick: when you see $\sqrt{1 \pm \cos \text{(something)}}$, think of $\cos 2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$.

Let's walk through it step by step.

1. Substitute $x = \cos 2\theta$.

   Since $x$ lies between $-\frac{1}{\sqrt{2}}$ and $1$, we have $\cos 2\theta$ in that range. This implies $2\theta$ is between $0$ and $\frac{3\pi}{4}$ (since $\cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}}$), so $\theta$ is between $0$ and $\frac{3\pi}{8}$. That's fine — we'll stay in the principal range where $\sin\theta$ and $\cos\theta$ are positive.

2. Simplify $\sqrt{1+x}$ and $\sqrt{1-x}$.

   Using $x = \cos 2\theta$:

Since $\theta$ is between $0$ and $\frac{3\pi}{8}$, $\cos\theta > 0$, so $|\cos\theta| = \cos\theta$. Thus $\sqrt{1+x} = \sqrt{2}\cos\theta$.

Similarly,

For $\theta$ in $(0, \frac{3\pi}{8})$, $\sin\theta > 0$, so $\sqrt{1-x} = \sqrt{2}\sin\theta$.

3. Plug into the fraction. The numerator becomes:

The denominator becomes:

The $\sqrt{2}$ cancels, so the fraction inside the $\tan^{-1}$ is:

4. Rewrite using tangent. Divide numerator and denominator by $\cos\theta$ (which is non-zero here):

This is a classic form: $\frac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right)$. Why? Because $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, and with $A = \frac{\pi}{4}$, $\tan\frac{\pi}{4} = 1$, we get exactly $\frac{1 - \tan\theta}{1 + \tan\theta}$.

So the expression becomes:

5. Check the range to apply the inverse. We need $\frac{\pi}{4} - \theta$ to lie in the principal range of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$. …

Method: Trigonometric substitution for radical inverse-trig identities

Use this for identities with $\sqrt{1+x}$ and $\sqrt{1-x}$ inside $\tan^{-1}$, guided by the hint $x = \cos 2\theta$.

Steps

Step 1: Substitute $x = \cos 2\theta$.

Then the half-angle forms apply: $1 + x = 2\cos^2\theta$ and $1 - x = 2\sin^2\theta$, so $\sqrt{1+x} = \sqrt{2}\,|\cos\theta|$ and $\sqrt{1-x} = \sqrt{2}\,|\sin\theta|$. Fix the signs from the given domain.

Step 2: Simplify the fraction to a tangent of a shifted angle. …

Common Mistakes

Mistake 1: Writing $\sqrt{2\cos^2\theta} = \sqrt{2}\cos\theta$ without the absolute value.

Why it's wrong: strictly $\sqrt{2\cos^2\theta} = \sqrt{2}\,|\cos\theta|$; the sign must be justified from the domain. Correct approach: the domain $-\frac{1}{\sqrt{2}} \le x \le 1$ gives $\theta \in \left[0, \frac{3\pi}{8}\right]$, where $\cos\theta$ and $\sin\theta$ are both positive.

Mistake 2: Applying $\tan^{-1}(\tan\phi) = \phi$ without a range check. …