Question

Evaluate: \[ \cot^{-1}\!\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) \]

• A. \(\frac{x}{3}\)
• B. \(\frac{x}{4}\)
• C. \(1\)
• D. \(\frac{x}{2}\)

Hint:

Expressions with \(\sqrt{1 \pm \sin x}\) often simplify using half-angle identities: \(1 \pm \sin x = (\sin\frac{x}{2} \pm \cos\frac{x}{2})^2\).

Answer 1

The Correct Option is D

To evaluate the given expression \(\cot^{-1}\!\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right)\), we can simplify it step by step.

The given expression is: \[ \cot^{-1}\!\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right). \]

Let's simplify the fraction inside the inverse cotangent.

We observe:

1. Set \(a = \sqrt{1+\sin x}\) and \(b = \sqrt{1-\sin x}\).
2. Thus, the question becomes \(\cot^{-1}\left(\frac{a + b}{a - b}\right)\).

We use the identity:

\[ a^2 - b^2 = (1 + \sin x) - (1 - \sin x) = 2\sin x. \]

Therefore, \[a^2 - b^2 = 2\sin x.\] And the expression becomes \(\frac{a + b}{a - b} = \left(\frac{a + b}{\sqrt{2\sin x}}\right)^2\).

Notice that this can also be seen through the tangent difference identity that relates \(\tan(A) = \frac{\sin A}{\cos A}\) . Here:

1. Let \(A = \frac{x}{2}\). Then \(\sin A = \sin \frac{x}{2}\) and \(\cos A = \cos\frac{x}{2}\).

Hence, the expression becomes \[ \cot^{-1}\left(\cot \frac{x}{2}\right), \] which simplifies to \(\frac{x}{2}\).

Thus, the correct answer is \(\frac{x}{2}\).