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

Substitute $x^2=\cos 2\theta$ (valid since $|x|\le1\Rightarrow x^2\in[0,1]$), which gives $2\theta\in\left[0,\frac{\pi}{2}\right]$, so $\theta\in\left[0,\frac{\pi}{4}\right]$ and $\cos\theta,\sin\theta\ge0$.

Using $1+\cos2\theta=2\cos^2\theta$ and $1-\cos2\theta=2\sin^2\theta$:

So the fraction becomes

$\frac{\sqrt2\cos\theta+\sqrt2\sin\theta}{\sqrt2\cos\theta-\sqrt2\sin\theta}=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=\frac{1+\tan\theta}{1-\tan\theta}=\tan\left(\frac{\pi}{4}+\theta\right).$ …

The idea

The expression is defined only when both $1+x^2$ and $1-x^2$ are non-negative, i.e. $|x|\le1$, so $x^2\in[0,1]$. Seeing $\sqrt{1\pm x^2}$ with $x^2$ over $[0,1]$ suggests writing $x^2=\cos2\theta$; then the half-angle identities dissolve the roots.

Step 1 — Substitute

Let $x^2=\cos2\theta$. Since $x^2\in[0,1]$, we have $\cos2\theta\in[0,1]$, so $2\theta\in\left[0,\frac{\pi}{2}\right]$ and $\theta\in\left[0,\frac{\pi}{4}\right]$. On this interval $\cos\theta\ge0$ and $\sin\theta\ge0$.

Step 2 — Kill the square roots

Using $1+\cos2\theta=2\cos^2\theta$ and $1-\cos2\theta=2\sin^2\theta$,

the absolute values dropping because both $\cos\theta,\sin\theta$ are non-negative here.

Step 3 — Simplify the fraction

Divide top and bottom by $\cos\theta$:

Step 4 — Take the inverse tangent (range check) …

Method: The $x^2=\cos 2\theta$ substitution for $\sqrt{1\pm x^2}$ expressions

Whenever an inverse-trig expression contains both $\sqrt{1+x^2}$ and $\sqrt{1-x^2}$ (with $|x|\le1$), a $\cos 2\theta$ substitution turns the square roots into single trig terms.

Steps

Step 1: Substitute and fix the range.

Since $|x|\le1$ gives $x^2\in[0,1]$, set $x^2=\cos 2\theta$; then $2\theta\in\left[0,\tfrac{\pi}{2}\right]$, so $\theta\in\left[0,\tfrac{\pi}{4}\right]$ and $\cos\theta,\sin\theta\ge0$.

Step 2: Remove the roots with half-angle identities.

the absolute values dropping because both are non-negative on this $\theta$-interval.

Step 3: Simplify to a single tangent. …

Common Mistakes

Mistake 1: Choosing the substitution $x=\cos 2\theta$ instead of $x^2=\cos 2\theta$.

Why it's wrong: the roots contain $1\pm x^2$, so it is $x^2$ (which lies in $[0,1]$) that should equal $\cos 2\theta$; using $x$ mismatches the half-angle step. Correct approach: set $x^2=\cos 2\theta$, giving $\theta\in\left[0,\tfrac{\pi}{4}\right]$.

Mistake 2: Dropping the absolute values carelessly when simplifying the roots.

Why it's wrong: $\sqrt{2\cos^2\theta}=\sqrt2\,|\cos\theta|$; the modulus can only be removed after confirming the sign. Correct approach: because $\theta\in\left[0,\tfrac{\pi}{4}\right]$ both $\cos\theta,\sin\theta\ge0$, so the roots become $\sqrt2\cos\theta$ and $\sqrt2\sin\theta$. …