Let `f(x) = {(sin (tan^(-1) x) + sin (cot^(-1) x)}^2 - 1`, where `|x| gt 1` and `dy/dx = 1/2 d/dx (sin^(-1) f(x))`. If `y(sqrt3) = pi/6` then `y( -sqrt3)`

Options

A. `-(pi)/(6)`
B. `(2pi)/(3)`
C. `(pi)/(3)`
D. `(5pi)/(6)`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Define the function \( f(x) \)
We start with the function given in the problem:
\[ f(x) = \left( \sin(\tan^{-1} x) + \sin(\cot^{-1} x) \right)^2 - 1 \]
where \( |x| > 1 \).

### Step 2: Simplify \( \sin(\cot^{-1} x) \)
Using the identity \( \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \), we can rewrite:
\[ \sin(\cot^{-1} x) = \sin\left(\frac{\pi}{2} - \tan^{-1} x\right) = \cos(\tan^{-1} x) \]

### Step 3: Substitute into \( f(x) \)
Now substituting back into \( f(x) \):
\[ f(x) = \left( \sin(\tan^{-1} x) + \cos(\tan^{-1} x) \right)^2 - 1 \]

### Step 4: Use the Pythagorean identity
Recall that \( \sin^2(\theta) + \cos^2(\theta) = 1 \). Therefore:
\[ \sin(\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}} \quad \text{and} \quad \cos(\tan^{-1} x) = \frac{1}{\sqrt{1+x^2}} \]
Thus:
\[ \sin(\tan^{-1} x) + \cos(\tan^{-1} x) = \frac{x + 1}{\sqrt{1+x^2}} \]

### Step 5: Substitute into \( f(x) \)
Now we can substitute this into \( f(x) \):
\[ f(x) = \left( \frac{x + 1}{\sqrt{1+x^2}} \right)^2 - 1 \]
\[ = \frac{(x + 1)^2}{1 + x^2} - 1 \]
\[ = \frac{(x + 1)^2 - (1 + x^2)}{1 + x^2} \]
\[ = \frac{x^2 + 2x + 1 - 1 - x^2}{1 + x^2} = \frac{2x}{1 + x^2} \]

### Step 6: Find \( \frac{dy}{dx} \)
From the problem, we have:
\[ \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx}(\sin^{-1} f(x)) \]
Using the chain rule:
\[ \frac{d}{dx}(\sin^{-1} f(x)) = \frac{1}{\sqrt{1 - f(x)^2}} \cdot f'(x) \]

### Step 7: Calculate \( f'(x) \)
We need to find \( f'(x) \):
\[ f(x) = \frac{2x}{1 + x^2} \]
Using the quotient rule:
\[ f'(x) = \frac{(1 + x^2)(2) - 2x(2x)}{(1 + x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2} \]

### Step 8: Substitute \( f(x) \) and \( f'(x) \) into \( \frac{dy}{dx} \)
Now substituting \( f(x) \) and \( f'(x) \) into \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} \cdot \frac{2(1 - x^2)}{(1 + x^2)^2} \]

### Step 9: Integrate to find \( y \)
Integrating \( \frac{dy}{dx} \) gives:
\[ y = \frac{1}{2} \sin^{-1} f(x) + C \]

### Step 10: Use the initial condition
Given \( y(\sqrt{3}) = \frac{\pi}{6} \):
\[ \frac{1}{2} \sin^{-1} f(\sqrt{3}) + C = \frac{\pi}{6} \]
Calculating \( f(\sqrt{3}) \):
\[ f(\sqrt{3}) = \frac{2\sqrt{3}}{1 + 3} = \frac{\sqrt{3}}{2} \]
Thus:
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \]
So:
\[ \frac{1}{2} \cdot \frac{\pi}{3} + C = \frac{\pi}{6} \]
This implies \( C = 0 \).

### Step 11: Find \( y(-\sqrt{3}) \)
Now, we need to find \( y(-\sqrt{3}) \):
\[ y(-\sqrt{3}) = \frac{1}{2} \sin^{-1} f(-\sqrt{3}) \]
Calculating \( f(-\sqrt{3}) \):
\[ f(-\sqrt{3}) = \frac{-2\sqrt{3}}{1 + 3} = -\frac{\sqrt{3}}{2} \]
Thus:
\[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \]
So:
\[ y(-\sqrt{3}) = \frac{1}{2} \left(-\frac{\pi}{3}\right) = -\frac{\pi}{6} \]

### Final Answer
The value of \( y(-\sqrt{3}) \) is:
\[ \boxed{-\frac{\pi}{6}} \]