If `cot(cos^(-1)x)=sec{tan^(-1)((a)/sqrt(b^(2)-a^(2)))}` then x equals

Options

A. `(b)/sqr(2b^(2)-a^(2))`
B. `(a)/sqr(2b^(2)-a^(2))`
C. `sqrt(b^(2)-a^(2))/(a)`
D. `b/sqrt(2b^2-a^2)`

Explanation

To solve the equation \( \cot(\cos^{-1} x) = \sec\left(\tan^{-1}\left(\frac{a}{\sqrt{b^2 - a^2}}\right)\right) \), we will follow these steps:

### Step 1: Define the angle for \( \cos^{-1} x \)
Let \( \theta = \cos^{-1} x \). Then, from the definition of cosine:
\[ \cos \theta = x \]
This implies that in a right triangle, the adjacent side is \( x \) and the hypotenuse is \( 1 \).

### Step 2: Find the opposite side
Using the Pythagorean theorem:
\[ \text{Opposite side} = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2} \]

### Step 3: Find \( \cot(\theta) \)
The cotangent of the angle \( \theta \) is given by:
\[ \cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{x}{\sqrt{1 - x^2}} \]

### Step 4: Define the angle for \( \tan^{-1} \left(\frac{a}{\sqrt{b^2 - a^2}}\right) \)
Let \( \phi = \tan^{-1}\left(\frac{a}{\sqrt{b^2 - a^2}}\right) \). Then:
\[ \tan \phi = \frac{a}{\sqrt{b^2 - a^2}} \]

### Step 5: Find \( \sec(\phi) \)
Using the definition of secant:
\[ \sec \phi = \frac{1}{\cos \phi} \]
To find \( \cos \phi \), we can use the identity:
\[ \sec^2 \phi = 1 + \tan^2 \phi \implies \sec^2 \phi = 1 + \left(\frac{a}{\sqrt{b^2 - a^2}}\right)^2 = 1 + \frac{a^2}{b^2 - a^2} \]
This simplifies to:
\[ \sec^2 \phi = \frac{b^2}{b^2 - a^2} \]
Thus,
\[ \sec \phi = \frac{b}{\sqrt{b^2 - a^2}} \]

### Step 6: Set the two expressions equal
Now we have:
\[ \cot(\cos^{-1} x) = \sec\left(\tan^{-1}\left(\frac{a}{\sqrt{b^2 - a^2}}\right)\right) \]
This gives us:
\[ \frac{x}{\sqrt{1 - x^2}} = \frac{b}{\sqrt{b^2 - a^2}} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ x \sqrt{b^2 - a^2} = b \sqrt{1 - x^2} \]
Squaring both sides results in:
\[ x^2 (b^2 - a^2) = b^2 (1 - x^2) \]
Expanding and rearranging yields:
\[ x^2 b^2 - x^2 a^2 = b^2 - b^2 x^2 \]
Combining like terms:
\[ x^2 (b^2 + a^2) = b^2 \]

### Step 8: Solve for \( x^2 \)
Thus, we have:
\[ x^2 = \frac{b^2}{b^2 + a^2} \]
Taking the square root gives:
\[ x = \frac{b}{\sqrt{b^2 + a^2}} \]

### Final Answer
So the value of \( x \) is:
\[ \boxed{\frac{b}{\sqrt{b^2 + a^2}}} \]