$ an^{-1} \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} ight) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x$, $-\frac{1}{\sqrt{2}} \le x \le 1$ [Hint: Put $x = \cos 2 heta$] Solve the following equations:

Question

$	an^{-1} \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}ight) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x$, $-\frac{1}{\sqrt{2}} \le x \le 1$ [Hint: Put $x = \cos 2	heta$] Solve the following equations:

Accepted Answer

> [!TLDR]
> The identity simplifies a nested radical expression into a neat inverse-trig form by substituting $x = \cos 2	heta$, using the half-angle formulas, and recognizing the standard inverse tangent identity. The final result is $\frac{\pi}{4} - \frac{1}{2}\cos^{-1} x$.

We need to show that for $-\frac{1}{\sqrt{2}} \le x \le 1$,
$$
	an^{-1} \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}ight) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x.
$$

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

Let's walk through it step by step.

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

2. **Simplify $\sqrt{1+x}$ and $\sqrt{1-x}$.**  
   Using $x = \cos 2	heta$:
   $$
   \sqrt{1+x} = \sqrt{1 + \cos 2	heta} = \sqrt{2\cos^2	heta} = \sqrt{2}\,|\cos	heta|.
   $$
   Since $	heta$ is between $0$ and $\frac{3\pi}{8}$, $\cos	heta > 0$, so $|\cos	heta| = \cos	heta$. Thus $\sqrt{1+x} = \sqrt{2}\cos	heta$.

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

3. **Plug into the fraction.**  
   The numerator becomes:
   $$
   \sqrt{1+x} - \sqrt{1-x} = \sqrt{2}\cos	heta - \sqrt{2}\sin	heta = \sqrt{2}(\cos	heta - \sin	heta).
   $$
   The denominator becomes:
   $$
   \sqrt{1+x} + \sqrt{

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 understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

Method: Trigonometric substitution for radical inverse-trig identities

Steps

Common Mistakes

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

Applying it to this problem

🎓Unlock everything free for 14 days

Unlock everything free for 14 days