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

Question

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

Accepted Answer

> [!TLDR]
> The substitution $x^2=\cos2	heta$ collapses the square roots into $\sqrt2\cos	heta$ and $\sqrt2\sin	heta$; the fraction becomes $	an\left(\frac{\pi}{4}+	hetaight)$, and since $\frac{\pi}{4}+	heta$ stays in the arctan principal range, the identity equals $\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^2$.

## 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 $x^2=\cos2	heta$; then the half-angle identities dissolve the roots.

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

## Step 2 — Kill the square roots
Using $1+\cos2	heta=2\cos^2	heta$ and $1-\cos2	heta=2\sin^2	heta$,
$$\sqrt{1+x^2}=\sqrt{2\cos^2	heta}=\sqrt2\,\cos	heta,\qquad\sqrt{1-x^2}=\sqrt{2\sin^2	heta}=\sqrt2\,\sin	heta,$$
the absolute values dropping because both $\cos	heta,\sin	heta$ are non-negative here.

## Step 3 — Simplify the fraction
$$\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\sqrt2\cos	heta+\sqrt2\sin	heta}{\sqrt2\cos	heta-\sqrt2\sin	heta}=\frac{\cos	heta+\sin	heta}{\cos	heta-\sin	heta}.$$
Divide top and bottom by $\cos	heta$:
$$\frac{1+	an	heta}{1-	an	heta}=\frac{	an\frac{\pi}{4}+	an	heta}{1-	an\frac{\pi}{4}	an	heta}=	an\left(\frac{\pi}{4}+	hetaight).$$

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

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

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

The idea

Step 1 — Substitute

Step 2 — Kill the square roots

Step 3 — Simplify the fraction

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

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

Steps

Common Mistakes

The core idea: the range of $\tan^{-1}$ and the product condition

Applying it to this problem

Q.Prove that tan⁡−1(1+x2+1−x21+x2−1−x2)=π4+12cos⁡−1x2\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^2tan−1(1+x2​−1−x2​1+x2​+1−x2​​)=4π​+21​cos−1x2.