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

Explanation

To solve the problem step by step, we start with the given function and differential equation.

### Step 1: Define the function and the differential equation
We have:
\[ f(x) = \left( \sin(\tan^{-1} x) + \sin(\cot^{-1} x) \right)^2 - 1 \]
and
\[ \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (\sin^{-1} f(x)) \]

### Step 2: Simplify \( f(x) \)
We know that:
- \( \tan^{-1} x = \theta \) implies \( \sin(\theta) = \frac{x}{\sqrt{1+x^2}} \)
- \( \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \) implies \( \sin(\cot^{-1} x) = \cos(\tan^{-1} x) = \frac{1}{\sqrt{1+x^2}} \)

Thus,
\[ f(x) = \left( \frac{x}{\sqrt{1+x^2}} + \frac{1}{\sqrt{1+x^2}} \right)^2 - 1 \]
\[ = \left( \frac{x + 1}{\sqrt{1+x^2}} \right)^2 - 1 \]
\[ = \frac{(x + 1)^2}{1+x^2} - 1 \]
\[ = \frac{(x + 1)^2 - (1 + x^2)}{1+x^2} \]
\[ = \frac{x^2 + 2x + 1 - 1 - x^2}{1+x^2} \]
\[ = \frac{2x}{1+x^2} \]

### Step 3: Substitute \( f(x) \) into the differential equation
Now substituting \( f(x) \):
\[ \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} \left( \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right) \]

### Step 4: Integrate both sides
Integrating both sides with respect to \( x \):
\[ y = \frac{1}{2} \sin^{-1} \left( \frac{2x}{1+x^2} \right) + C \]

### Step 5: Use the initial condition to find \( C \)
We know that \( y(\sqrt{3}) = \frac{\pi}{6} \).
Substituting \( x = \sqrt{3} \):
\[ y(\sqrt{3}) = \frac{1}{2} \sin^{-1} \left( \frac{2\sqrt{3}}{1 + 3} \right) + C \]
\[ = \frac{1}{2} \sin^{-1} \left( \frac{2\sqrt{3}}{4} \right) + C \]
\[ = \frac{1}{2} \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) + C \]
\[ = \frac{1}{2} \cdot \frac{\pi}{3} + C \]
\[ = \frac{\pi}{6} + C \]

Setting this equal to \( \frac{\pi}{6} \):
\[ \frac{\pi}{6} + C = \frac{\pi}{6} \]
\[ C = 0 \]

### Step 6: Final expression for \( y \)
Thus, the final expression for \( y \) is:
\[ y = \frac{1}{2} \sin^{-1} \left( \frac{2x}{1+x^2} \right) \]

### Step 7: Find \( y(-\sqrt{3}) \)
Now we need to find \( y(-\sqrt{3}) \):
\[ y(-\sqrt{3}) = \frac{1}{2} \sin^{-1} \left( \frac{2(-\sqrt{3})}{1 + 3} \right) \]
\[ = \frac{1}{2} \sin^{-1} \left( \frac{-2\sqrt{3}}{4} \right) \]
\[ = \frac{1}{2} \sin^{-1} \left( -\frac{\sqrt{3}}{2} \right) \]
\[ = \frac{1}{2} \left( -\frac{\pi}{3} \right) \]
\[ = -\frac{\pi}{6} \]

### Final Answer
Thus, \( y(-\sqrt{3}) = -\frac{\pi}{6} \).

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