If `x=sin(2tan^(-1)2), y = sin (1/2 tan^(-1)"4/3)`, then

Options

A. `x gt y and y^2 =1-x`
B. `x gt y`
C. `x gt y and y^2 =x`
D. `y^2=1+x`

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) using the given expressions:

1. **Finding \( x \):**
\[ x = \sin(2 \tan^{-1}(2)) \]
Let \( \theta = \tan^{-1}(2) \). Then, we have:
\[ \tan(\theta) = 2 \]
Using the double angle formula for sine:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
We need to find \( \sin(\theta) \) and \( \cos(\theta) \).

From \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{1} \), we can find the hypotenuse:
\[ \text{hypotenuse} = \sqrt{2^2 + 1^2} = \sqrt{5} \]
Thus:
\[ \sin(\theta) = \frac{2}{\sqrt{5}}, \quad \cos(\theta) = \frac{1}{\sqrt{5}} \]
Now substituting these into the double angle formula:
\[ x = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{2}{\sqrt{5}}\right) \left(\frac{1}{\sqrt{5}}\right) = \frac{4}{5} \]

2. **Finding \( y \):**
\[ y = \sin\left(\frac{1}{2} \tan^{-1}\left(\frac{4}{3}\right)\right) \]
Let \( \beta = \tan^{-1}\left(\frac{4}{3}\right) \). Then:
\[ \tan(\beta) = \frac{4}{3} \]
Again, we find the hypotenuse:
\[ \text{hypotenuse} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \]
Thus:
\[ \sin(\beta) = \frac{4}{5}, \quad \cos(\beta) = \frac{3}{5} \]
Using the half angle formula:
\[ y = \sin\left(\frac{\beta}{2}\right) = \sqrt{\frac{1 - \cos(\beta)}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

3. **Comparing \( x \) and \( y \):**
We have:
\[ x = \frac{4}{5}, \quad y = \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} \approx 0.447 \]
Since \( \frac{4}{5} = 0.8 \) and \( 0.8 > 0.447 \), we find that \( x > y \).

4. **Verifying the relationship:**
To show \( y^2 = 1 - x \):
\[ y^2 = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5} \]
And:
\[ 1 - x = 1 - \frac{4}{5} = \frac{1}{5} \]
Thus, \( y^2 = 1 - x \).

### Final Results:
- \( x = \frac{4}{5} \)
- \( y = \frac{1}{\sqrt{5}} \)
- \( y^2 = 1 - x \)