The expression `1/(sqrt(2)){(sincot^(- 1)costan^(- 1)t)/(costan^(- 1)sincot^(- 1)sqrt(2)t)}*{sqrt((1+2t^2)/(2+t^2))}`can take the value

Options

A. `1//2`
B. `-5`
C. `1`
D. `3//4`

Explanation

To solve the expression

\[ \frac{1}{\sqrt{2}} \cdot \frac{\sin(\cot^{-1}(\cos(\tan^{-1} t)))}{\cos(\tan^{-1}(\sin(\cot^{-1}(\sqrt{2}t))))} \cdot \sqrt{\frac{1 + 2t^2}{2 + t^2}}, \]

we will simplify the numerator and denominator step by step.

### Step 1: Simplifying the Numerator

Let \( \tan^{-1}(t) = \alpha \). Then, we have:

\[ t = \tan(\alpha). \]

From the definition of tangent, we can construct a right triangle where the opposite side is \( t \) and the adjacent side is \( 1 \). The hypotenuse \( h \) can be calculated as:

\[ h = \sqrt{t^2 + 1} = \sqrt{\tan^2(\alpha) + 1} = \sec(\alpha). \]

Now, we can find \( \sin(\cot^{-1}(\cos(\alpha))) \). Let \( \beta = \cot^{-1}(\cos(\alpha)) \). Then,

\[ \cot(\beta) = \cos(\alpha) = \frac{1}{\sec(\alpha)} = \frac{1}{h}. \]

In a right triangle, if \( \cot(\beta) = \frac{\text{adjacent}}{\text{opposite}} \), we can set the adjacent side to \( 1 \) and the opposite side to \( h \). Thus, the hypotenuse \( H \) can be calculated as:

\[ H = \sqrt{1^2 + h^2} = \sqrt{1 + (t^2 + 1)} = \sqrt{t^2 + 2}. \]

Now, we can find \( \sin(\beta) \):

\[ \sin(\beta) = \frac{h}{H} = \frac{\sqrt{t^2 + 1}}{\sqrt{t^2 + 2}}. \]

### Step 2: Simplifying the Denominator

Now, consider the denominator:

Let \( \cot^{-1}(\sqrt{2}t) = \theta \). Then,

\[ \cot(\theta) = \sqrt{2}t. \]

Using a similar triangle approach, we can find \( \sin(\theta) \):

\[ \sin(\theta) = \frac{1}{\sqrt{(\sqrt{2}t)^2 + 1}} = \frac{1}{\sqrt{2t^2 + 1}}. \]

Now, we need to find \( \cos(\tan^{-1}(\sin(\theta))) \). Let \( \gamma = \tan^{-1}(\sin(\theta)) \). Then,

\[ \tan(\gamma) = \sin(\theta) = \frac{1}{\sqrt{2t^2 + 1}}. \]

Using the triangle definition again, we find:

\[ \cos(\gamma) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{2t^2 + 1}}{\sqrt{(2t^2 + 1) + 1}} = \frac{\sqrt{2t^2 + 1}}{\sqrt{2t^2 + 2}} = \frac{\sqrt{2t^2 + 1}}{\sqrt{2(t^2 + 1)}}. \]

### Step 3: Putting It All Together

Now we can substitute back into our original expression:

\[ \frac{1}{\sqrt{2}} \cdot \frac{\frac{\sqrt{t^2 + 1}}{\sqrt{t^2 + 2}}}{\frac{\sqrt{2t^2 + 1}}{\sqrt{2(t^2 + 1)}}} \cdot \sqrt{\frac{1 + 2t^2}{2 + t^2}}. \]

This simplifies to:

\[ \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{t^2 + 1} \cdot \sqrt{2(t^2 + 1)}}{\sqrt{t^2 + 2} \cdot \sqrt{2t^2 + 1}} \cdot \sqrt{\frac{1 + 2t^2}{2 + t^2}}. \]

### Step 4: Final Simplification

After simplifying all terms, we can see that the expression can take values depending on \( t \). The minimum value occurs when \( t^2 + 2 \) is minimized, which is 2. Thus, the expression can take values less than 1.

### Conclusion

The expression can take values in the range \( \left( \frac{1}{2}, 1 \right) \).