Question

$\frac{d}{dx} \left[ \cos^2 \left( \cot^{-1} \sqrt{\frac{2 + x}{2 - x}} \right) \right]$ is:}

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $\frac{1}{4}$

Answer 1

The Correct Option is D

1. Understand the problem:

We need to differentiate the expression \( \cos^2 \left( \cot^{-1} \sqrt{\frac{2+x}{2-x}} \right) \) with respect to \( x \).

2. Simplify the expression:

Let \( \theta = \cot^{-1} \sqrt{\frac{2+x}{2-x}} \). Then, \( \cot \theta = \sqrt{\frac{2+x}{2-x}} \).

Construct a right triangle with adjacent side \( \sqrt{2+x} \) and opposite side \( \sqrt{2-x} \). The hypotenuse is:

\[ \sqrt{(\sqrt{2+x})^2 + (\sqrt{2-x})^2} = \sqrt{2+x + 2-x} = \sqrt{4} = 2 \]

Thus, \( \cos \theta = \frac{\sqrt{2+x}}{2} \), and:

\[ \cos^2 \theta = \left( \frac{\sqrt{2+x}}{2} \right)^2 = \frac{2+x}{4} \]

3. Differentiate with respect to \( x \):

Now, the expression simplifies to \( \frac{2+x}{4} \). Differentiate:

\[ \frac{d}{dx} \left( \frac{2+x}{4} \right) = \frac{1}{4} \]

Correct Answer: (D) \( \frac{1}{4} \)

Answer 2

1. Let's simplify the expression first

Let $ \theta = \cot^{-1}\left(\frac{\sqrt{2+x}}{\sqrt{2-x}}\right) $.

Then, $ \cot(\theta) = \frac{\sqrt{2+x}}{\sqrt{2-x}} $.

Consider a right-angled triangle where the adjacent side is $ \sqrt{2+x} $ and the opposite side is $ \sqrt{2-x} $.

Then the hypotenuse is:

\[ \sqrt{\left(\sqrt{2+x}\right)^2 + \left(\sqrt{2-x}\right)^2} = \sqrt{(2+x) + (2-x)} = \sqrt{4} = 2 \]

Therefore, $ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{2+x}}{2} $.

So, $ \cos^2\left(\cot^{-1}\left(\frac{\sqrt{2+x}}{\sqrt{2-x}}\right)\right) = \cos^2(\theta) = \left(\frac{\sqrt{2+x}}{2}\right)^2 = \frac{2+x}{4} $.

2. Now differentiate:

\[ \frac{d}{dx} \left[\cos^2\left(\cot^{-1}\left(\frac{\sqrt{2+x}}{\sqrt{2-x}}\right)\right)\right] = \frac{d}{dx} \left[\frac{2+x}{4}\right] = \frac{1}{4} \]

Therefore, the derivative is $ \frac{1}{4} $.