Let `x_(1) " and " x_(2) ( x_(1) gt x_(2))` be roots of the equation `sin^(-1) ( cos ( tan^(-1)( cosec ( cot^(-1)x)))) = pi/6 `, then

Options

A. ` sin ^(-1). 1/x^(1) + cos ^(-1). 1/x_(2) = pi`
B. `sin^(-1) ( 1/x_(1)) + cos ^(-1) ( 1/x_(2)) = 0 `
C. ` sin ^(-1) . 1/x_(1) + sin ^(-1) ( 1/x_(2)) = 0 `
D. ` cos^(-1) ( 1/x_(1)) + cos^(-1) ( 1/x_(2)) = pi `

Answer

To solve the equation \( \sin^{-1} \left( \cos \left( \tan^{-1} \left( \csc \left( \cot^{-1} x \right) \right) \right) \right) = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Substitute \( A = \cot^{-1} x \)
Let \( A = \cot^{-1} x \). Then, we have:
\[ \cot A = x \]
This means we can construct a right triangle where the adjacent side is 1 and the opposite side is \( x \). By the Pythagorean theorem, the hypotenuse will be:
\[ \text{Hypotenuse} = \sqrt{x^2 + 1} \]

### Step 2: Find \( \csc A \)
From the triangle, we can find \( \csc A \):
\[ \csc A = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{\sqrt{x^2 + 1}}{1} = \sqrt{x^2 + 1} \]

### Step 3: Substitute \( \csc A \) into the equation
Now, we substitute \( \csc A \) into the original equation:
\[ \sin^{-1} \left( \cos \left( \tan^{-1} \left( \sqrt{x^2 + 1} \right) \right) \right) = \frac{\pi}{6} \]

### Step 4: Let \( V = \tan^{-1} \left( \sqrt{x^2 + 1} \right) \)
Let \( V = \tan^{-1} \left( \sqrt{x^2 + 1} \right) \). Then:
\[ \tan V = \sqrt{x^2 + 1} \]
Constructing a right triangle for angle \( V \), we have:
- Opposite = \( \sqrt{x^2 + 1} \)
- Adjacent = 1
- Hypotenuse = \( \sqrt{(\sqrt{x^2 + 1})^2 + 1^2} = \sqrt{x^2 + 2} \)

### Step 5: Find \( \cos V \)
From the triangle, we can find \( \cos V \):
\[ \cos V = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{x^2 + 2}} \]

### Step 6: Substitute \( \cos V \) into the equation
Now, we substitute \( \cos V \) back into the equation:
\[ \sin^{-1} \left( \frac{1}{\sqrt{x^2 + 2}} \right) = \frac{\pi}{6} \]

### Step 7: Solve the equation
Taking the sine of both sides:
\[ \frac{1}{\sqrt{x^2 + 2}} = \sin \frac{\pi}{6} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2 = \sqrt{x^2 + 2} \]
Squaring both sides:
\[ 4 = x^2 + 2 \]
Thus, we have:
\[ x^2 = 2 \]
Taking the square root:
\[ x = \pm \sqrt{2} \]

### Step 8: Identify the roots
Given \( x_1 > x_2 \), we have:
\[ x_1 = \sqrt{2}, \quad x_2 = -\sqrt{2} \]

### Conclusion
The roots of the equation are \( x_1 = \sqrt{2} \) and \( x_2 = -\sqrt{2} \).

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