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

Options

A. `x=y^(2)`
B. `y^(2)=1-x`
C. `x^(2)=(y)/(2)`
D. `y^(2)=1+x`

Answer

To solve the problem step by step, we will calculate the values of \( x \) and \( y \) as defined in the question.

### Step 1: Calculate \( x = \sin(2 \tan^{-1}(2)) \)

We can use the double angle formula for sine:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Let \( \theta = \tan^{-1}(2) \). Then, we have:
\[ \tan(\theta) = 2 \implies \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{1} \]
Using the Pythagorean theorem, we find the hypotenuse:
\[ \text{hypotenuse} = \sqrt{2^2 + 1^2} = \sqrt{5} \]
Now, we can find \( \sin(\theta) \) and \( \cos(\theta) \):
\[ \sin(\theta) = \frac{2}{\sqrt{5}}, \quad \cos(\theta) = \frac{1}{\sqrt{5}} \]
Now substituting back into the sine double angle formula:
\[ x = \sin(2 \tan^{-1}(2)) = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{2}{\sqrt{5}}\right) \left(\frac{1}{\sqrt{5}}\right) = \frac{4}{5} \]

### Step 2: Calculate \( y = \sin\left(\frac{1}{2} \tan^{-1}\left(\frac{4}{3}\right)\right) \)

Let \( \phi = \tan^{-1}\left(\frac{4}{3}\right) \). Then, we have:
\[ \tan(\phi) = \frac{4}{3} \implies \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3} \]
Using the Pythagorean theorem again:
\[ \text{hypotenuse} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \]
Now we can find \( \sin(\phi) \) and \( \cos(\phi) \):
\[ \sin(\phi) = \frac{4}{5}, \quad \cos(\phi) = \frac{3}{5} \]
Now we need to find \( \sin\left(\frac{1}{2} \phi\right) \) using the half-angle formula:
\[ \sin\left(\frac{\phi}{2}\right) = \sqrt{\frac{1 - \cos(\phi)}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]
Thus, \( y = \frac{1}{\sqrt{5}} \).

### Step 3: Relate \( x \) and \( y \)

From the calculations:
\[ x = \frac{4}{5}, \quad y = \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} \]
Now, we can square \( y \):
\[ y^2 = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5} \]
Also, we can express \( x \) in terms of \( y \):
\[ 1 - x = 1 - \frac{4}{5} = \frac{1}{5} \]
Thus, we have:
\[ y^2 = 1 - x \]

### Final Conclusion

The relationship we derived is:
\[ y^2 = 1 - x \]
This means that the values of \( x \) and \( y \) are related as required.