Q.Differentiate w.r.t. $x$: $\tan^{-1}\left(\sqrt{\dfrac{1 - \cos x}{1 + \cos x}}\right),\ -\dfrac{\pi}{4} < x < \dfrac{\pi}{4}$.

The key idea: For a composition like $\sin^{-1}(\sqrt{x-1})$, we need both the inner function ($\sqrt{x-1}$) to be defined and its output to lie within the domain of the outer function ($\sin^{-1}$). The inverse sine accepts inputs only between $-1$ and $1$, and the square root requires a non-negative radicand and produces non-negative outputs.

1. Domain of the square root $\sqrt{x-1}$ is defined only when the radicand is non-negative:

2. Range of the square root For $x\ge 1$, $\sqrt{x-1}$ takes values from $0$ to $\infty$. But the outer function $\sin^{-1}(t)$ is defined only for $t\in[-1,1]$. Since $\sqrt{x-1}\ge 0$, we only need to ensure

3. Solve the inequality Square both sides (both sides are non-negative, so squaring preserves the inequality):

Adding 1 throughout:

4. Combine with step 1 …

Step 1 — Identify the two constraints.

• For $\sqrt{\;\cdot\;}$ to be real: $(x-2)(x-5)\ge 0$.
• For the fraction to be defined (no division by zero): $\sqrt{(x-2)(x-5)}\ne 0$, i.e. $(x-2)(x-5)\ne 0$.

Combining, the single condition is the strict inequality:

This is the whole point of the question — the endpoints $x=2$ and $x=5$ are excluded precisely because the root sits downstairs.

Step 2 — Solve by sign analysis.

Critical points: $x=2$ and $x=5$. Test each interval:

We need the set of real $x$ for which $f(x)=\sqrt{\sin^{-1}(2x)+\frac{\pi}{6}}$ is defined. Two conditions must hold simultaneously:

1. The argument of $\sin^{-1}(2x)$ must lie in $[-1,1]$, because $\sin^{-1}(t)$ is only defined for $t\in[-1,1]$.
2. The quantity under the square root must be non‑negative: $\sin^{-1}(2x)+\frac{\pi}{6} \ge 0$.

Let’s work through these step by step.

1. Domain of $\sin^{-1}(2x)$ Since $\sin^{-1}(u)$ is defined for $u \in [-1,1]$, we require

So the “outer” bounds are $x\in\left[-\frac12,\frac12\right]$.

2. Square‑root condition We need $\sin^{-1}(2x) + \frac{\pi}{6} \ge 0$, i.e.

The inverse sine function $\sin^{-1}(t)$ is increasing on $[-1,1]$, with range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

The inequality $\sin^{-1}(2x) \ge -\frac{\pi}{6}$ means $2x$ must be at least the value whose sine is $-\frac{\pi}{6}$.

Since $\sin\left(-\frac{\pi}{6}\right) = -\frac12$, we have

3. Combine both conditions …

The key here is to see what $g(f(x))$ actually looks like. When you have a composition like this, don't just leave it as a mess — simplify it. The function $f(x) = \sin 2x + \cos 2x$ can be rewritten as $\sqrt{2} \sin\left(2x + \frac{\pi}{4}\right)$, but even more useful is to square it when plugging into $g$.

Since $g(x) = x^2 - 1$, we get $g(f(x)) = (\sin 2x + \cos 2x)^2 - 1$. Expand that square: $\sin^2 2x + \cos^2 2x + 2\sin 2x \cos 2x - 1$. The $\sin^2 + \cos^2$ gives $1$, so $1 + \sin 4x - 1 = \sin 4x$. That's a huge simplification.

So the problem reduces to: for which domain is $\sin 4x$ invertible? A function is invertible only if it is one-to-one (bijective) on that domain. For a sine function, that means the domain must be contained within an interval of length $\frac{\pi}{2}$ where sine is strictly monotonic — but here the argument is $4x$, so the period is compressed.

1. Understand the period and monotonicity of $\sin 4x$

   The function $\sin 4x$ has period $\frac{2\pi}{4} = \frac{\pi}{2}$. It is one-to-one on any interval of length $\frac{\pi}{4}$ where it is strictly increasing or strictly decreasing. For example, $\sin \theta$ is one-to-one on $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so $\sin 4x$ is one-to-one when $4x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, i.e., $x \in [-\frac{\pi}{8}, \frac{\pi}{8}]$.

2. Check each option

   • Option (A): $[-\frac{\pi}{8}, \frac{\pi}{8}]$ — length $\frac{\pi}{4}$, exactly the interval where $\sin 4x$ goes from $-1$ to $1$ monotonically. This works.
   • Option (B): $[-\frac{\pi}{2}, \frac{\pi}{2}]$ — length $\pi$, which is $4$ times the monotonic interval. $\sin 4x$ will go through multiple cycles here, so it's not one-to-one. …