Question

The value of
\(\cot^{-1}[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}]\) where x ∈ \((0,\frac{\pi}{4})\) is

• A. \(\pi-\frac{x}{3}\)
• B. \(\frac{x}{2}\)
• C. \(\pi-\frac{x}{2}\)
• D. \(\frac{x}{2}-\pi\)

Hint:

In trigonometric simplifications, look for standard identities and use the conjugate to simplify fractions involving square roots. This technique is especially useful when dealing with expressions inside inverse trigonometric functions.

Answer 1

The Correct Option is C

The correct answer is (C) : \(\pi-\frac{x}{2}\).

Answer 2

The correct answer is: (C) \( \pi - \frac{x}{2} \) .

We are asked to find the value of:

\( \cot^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right] \) where \( x \in \left(0, \frac{\pi}{4}\right) \).

To solve this, we start by simplifying the given expression inside the inverse cotangent function. Let's focus on simplifying the fraction:

\( \frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}} \)

This is a standard form that can be simplified by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of \( \sqrt{1-\sin x} - \sqrt{1+\sin x} \) is \( \sqrt{1-\sin x} + \sqrt{1+\sin x} \). This simplification leads to the following:

\( \frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})^2}{(\sqrt{1-\sin x})^2 - (\sqrt{1+\sin x})^2} \)

Using the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \), we get:

\( \frac{(1-\sin x) + 2\sqrt{(1-\sin x)(1+\sin x)} + (1+\sin x)}{(1-\sin x) - (1+\sin x)} \)

Simplifying further, we end up with an expression involving \( \tan \left(\frac{x}{2}\right) \), leading to the final answer:

\( \pi - \frac{x}{2} \)

Therefore, the correct answer is (C) \( \pi - \frac{x}{2} \) .