Question

If \(x\in\left(0,\dfrac{\pi}{4}\right)\), then prove that \(\cot^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\dfrac{x}{2}\).

Hint:

Write \(1\pm\sin x\) as \((\cos\tfrac{x}{2}\pm\sin\tfrac{x}{2})^2\) using half angle formulas.

Answer 1

Step 1: Write what has to be shown:
We must prove that the given inverse cotangent expression equals \(x/2\) for \(x\in(0,\pi/4)\).
The key idea is to write \(1+\sin x\) and \(1-\sin x\) as perfect squares using half angle values.

Step 2: Convert to half angle form:
Using \(\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\) and \(1=\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}\), we get two perfect squares.
\[ 1+\sin x = \left(\cos\tfrac{x}{2}+\sin\tfrac{x}{2}\right)^2, \qquad 1-\sin x = \left(\cos\tfrac{x}{2}-\sin\tfrac{x}{2}\right)^2 \]
Since \(x\in(0,\pi/4)\), we have \(x/2\in(0,\pi/8)\), so \(\cos\tfrac{x}{2} > \sin\tfrac{x}{2} > 0\).
Both \(\cos\tfrac{x}{2}+\sin\tfrac{x}{2}\) and \(\cos\tfrac{x}{2}-\sin\tfrac{x}{2}\) are positive, so square roots give these values directly, no sign issue.

Step 3: Take square roots and simplify the ratio:
\[ \sqrt{1+\sin x}=\cos\tfrac{x}{2}+\sin\tfrac{x}{2}, \qquad \sqrt{1-\sin x}=\cos\tfrac{x}{2}-\sin\tfrac{x}{2} \]
Adding and subtracting these two results:
\[ \sqrt{1+\sin x}+\sqrt{1-\sin x}=2\cos\tfrac{x}{2}, \qquad \sqrt{1+\sin x}-\sqrt{1-\sin x}=2\sin\tfrac{x}{2} \]
So the ratio inside the inverse cotangent becomes:
\[ \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\frac{2\cos\tfrac{x}{2}}{2\sin\tfrac{x}{2}}=\cot\tfrac{x}{2} \]

Step 4: Apply the inverse function:
Since \(x/2\in(0,\pi/8)\subset(0,\pi)\), which is exactly the principal range of \(\cot^{-1}\), we can invert directly.
\[ \cot^{-1}\left(\cot\tfrac{x}{2}\right)=\frac{x}{2} \]
This matches the right hand side of the identity, so the statement is true for every \(x\) in the given interval.

Final Answer:
Hence the identity \(\cot^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\dfrac{x}{2}\) is proved. \[ \boxed{\cot^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\dfrac{x}{2}} \]