Q.Differentiate the function $\cot^{-1} \left[\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right], 0 < x < \frac{\pi}{2}$ with respect to $x$.

Concept: Chain Rule — simplify the composite expression first, then differentiate.

Step 1: Using $1\pm\sin x=\left(\sin\tfrac x2\pm\cos\tfrac x2\right)^2$, the bracket reduces to $\cot\tfrac x2$.

Step 2: So $y = \cot^{-1}(\cot\tfrac x2) = \dfrac x2$ since $\tfrac x2 \in (0,\tfrac\pi4) \subset (0,\pi)$. …

Step 1 — Simplify $1\pm\sin x$ as perfect squares

Step 2 — Take square roots with the correct sign

For $0<x<\tfrac\pi2$, $0<\tfrac x2<\tfrac\pi4$, so $\cos\tfrac x2>\sin\tfrac x2>0$. Since a square root is non-negative,

Step 3 — Simplify the fraction

Step 4 — Apply $\cot^{-1}$

Since $\tfrac x2\in\left(0,\tfrac\pi4\right)\subset(0,\pi)$, the principal branch of $\cot^{-1}$,

Step 5 — Differentiate …

Method: Simplify an Inverse-Trig Expression via Identities, THEN Differentiate

Some inverse-trig problems bury a very simple expression inside a complicated-looking radical fraction. Simplify using half-angle identities first to find the expression's true value as a function of $x$ — but remember that simplifying is only the FIRST half of the problem when the question asks for a derivative; differentiating the simplified result is the second, required half.

Steps

Step 1: Rewrite $1\pm\sin x$ (or similar) as perfect squares using half-angle identities

$1+\sin x=\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2$ and $1-\sin x=\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)^2$.

Step 2: Take square roots carefully, using the given domain to fix the correct sign

Step 3: Simplify the resulting fraction to a recognisable trig ratio (e.g. $\cot\frac{x}{2}$), and apply the inverse function

Confirm the result lies in the inverse function's principal branch before writing $\cot^{-1}(\cot\theta)=\theta$. …

Common Mistakes

Mistake 1: Using the wrong sign when taking the square root of $\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)^2$.

Why it's wrong: a square root is always non-negative, so $\sqrt{(\cdot)^2}=|\cdot|$ — for $0<x<\frac{\pi}{2}$, $\cos\frac{x}{2}>\sin\frac{x}{2}$, so the correct simplification is $\cos\frac{x}{2}-\sin\frac{x}{2}$, not $\sin\frac{x}{2}-\cos\frac{x}{2}$. Correct approach: determine the sign of the quantity inside the absolute value using the given domain before removing the square root.

Mistake 2: Stopping at the simplified value $\frac{x}{2}$ and treating it as the final answer to a differentiation question. …