`(1/y^2 \ ((cos(tan^(-1) y) + y sin(tan^(-1) y))/(cot(sin^(-1) y) + tan(sin^(-1) y)) )^2 + y^4)^(1//2)` takes value

Answer

To solve the expression

\[ \left( \frac{1}{y^2} \left( \frac{\cos(\tan^{-1} y) + y \sin(\tan^{-1} y)}{\cot(\sin^{-1} y) + \tan(\sin^{-1} y)} \right)^2 + y^4 \right)^{\frac{1}{2}}, \]

we will break it down step by step.

### Step 1: Simplify the expression inside the parentheses

We need to simplify

\[ \frac{\cos(\tan^{-1} y) + y \sin(\tan^{-1} y)}{\cot(\sin^{-1} y) + \tan(\sin^{-1} y)}. \]

### Step 2: Find \(\cos(\tan^{-1} y)\) and \(\sin(\tan^{-1} y)\)

Using the right triangle definition, if \(\tan^{-1} y = \theta\), then:

- Opposite side = \(y\)
- Adjacent side = \(1\)
- Hypotenuse = \(\sqrt{y^2 + 1}\)

Thus,

\[ \cos(\tan^{-1} y) = \frac{1}{\sqrt{y^2 + 1}}, \quad \sin(\tan^{-1} y) = \frac{y}{\sqrt{y^2 + 1}}. \]

### Step 3: Substitute into the expression

Now substituting these values into the numerator:

\[ \cos(\tan^{-1} y) + y \sin(\tan^{-1} y) = \frac{1}{\sqrt{y^2 + 1}} + y \cdot \frac{y}{\sqrt{y^2 + 1}} = \frac{1 + y^2}{\sqrt{y^2 + 1}}. \]

### Step 4: Find \(\cot(\sin^{-1} y)\) and \(\tan(\sin^{-1} y)\)

Using the right triangle definition, if \(\sin^{-1} y = \phi\), then:

- Opposite side = \(y\)
- Hypotenuse = \(1\)
- Base = \(\sqrt{1 - y^2}\)

Thus,

\[ \cot(\sin^{-1} y) = \frac{\text{Base}}{\text{Opposite}} = \frac{\sqrt{1 - y^2}}{y}, \quad \tan(\sin^{-1} y) = \frac{y}{\sqrt{1 - y^2}}. \]

### Step 5: Substitute into the denominator

Now substituting these values into the denominator:

\[ \cot(\sin^{-1} y) + \tan(\sin^{-1} y) = \frac{\sqrt{1 - y^2}}{y} + \frac{y}{\sqrt{1 - y^2}}. \]

### Step 6: Combine the fractions

To combine the fractions in the denominator:

\[ \frac{\sqrt{1 - y^2}}{y} + \frac{y}{\sqrt{1 - y^2}} = \frac{(1 - y^2) + y^2}{y \sqrt{1 - y^2}} = \frac{1}{y \sqrt{1 - y^2}}. \]

### Step 7: Substitute back into the main expression

Now we can substitute back into the main expression:

\[ \frac{\frac{1 + y^2}{\sqrt{y^2 + 1}}}{\frac{1}{y \sqrt{1 - y^2}}} = \frac{(1 + y^2) y \sqrt{1 - y^2}}{\sqrt{y^2 + 1}}. \]

### Step 8: Square the result

Now we square the entire expression:

\[ \left( \frac{(1 + y^2) y \sqrt{1 - y^2}}{\sqrt{y^2 + 1}} \right)^2 = \frac{(1 + y^2)^2 y^2 (1 - y^2)}{y^2 + 1}. \]

### Step 9: Add \(y^4\) and simplify

Now we add \(y^4\):

\[ \frac{(1 + y^2)^2 y^2 (1 - y^2)}{y^2 + 1} + y^4. \]

### Step 10: Factor and simplify

After simplification, we find:

\[ \frac{(1 + y^2)^2 y^2 (1 - y^2) + y^4 (y^2 + 1)}{y^2 + 1}. \]

### Step 11: Take the square root

Finally, we take the square root:

\[ \left( \frac{1}{y^2} \left( \text{result} \right) + y^4 \right)^{\frac{1}{2}}. \]

### Final Result

After simplification, we find that the expression evaluates to \(1\).