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

Options

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

Answer

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

\[ x = \sin(2 \tan^{-1}(3)) \]
\[ y = \sin\left(\frac{1}{2} \tan^{-1}\left(\frac{4}{3}\right)\right) \]

### Step 1: Calculate \( x \)

Let \( \theta = \tan^{-1}(3) \). Therefore, we have:

\[ x = \sin(2\theta) \]

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 \( \theta = \tan^{-1}(3) \), we know:

\[ \tan(\theta) = \frac{3}{1} = 3 \]

This can be represented in a right triangle where the opposite side is 3 and the adjacent side is 1. The hypotenuse \( h \) can be calculated using the Pythagorean theorem:

\[ h = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]

Now we can find \( \sin(\theta) \) and \( \cos(\theta) \):

\[ \sin(\theta) = \frac{3}{\sqrt{10}}, \quad \cos(\theta) = \frac{1}{\sqrt{10}} \]

Substituting these values into the double angle formula:

\[ x = 2 \sin(\theta) \cos(\theta) = 2 \cdot \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5} \]

### Step 2: Calculate \( y \)

Now, let \( \phi = \tan^{-1}\left(\frac{4}{3}\right) \). Therefore:

\[ y = \sin\left(\frac{1}{2} \phi\right) \]

Using the half angle formula for sine:

\[ \sin\left(\frac{1}{2} \phi\right) = \sqrt{\frac{1 - \cos(\phi)}{2}} \]

We need to find \( \cos(\phi) \). From \( \phi = \tan^{-1}\left(\frac{4}{3}\right) \):

\[ \tan(\phi) = \frac{4}{3} \]

This can be represented in a right triangle where the opposite side is 4 and the adjacent side is 3. The hypotenuse \( h \) can be calculated:

\[ h = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Now we can find \( \sin(\phi) \) and \( \cos(\phi) \):

\[ \sin(\phi) = \frac{4}{5}, \quad \cos(\phi) = \frac{3}{5} \]

Substituting \( \cos(\phi) \) into the half angle formula:

\[ y = \sqrt{\frac{1 - \cos(\phi)}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

### Step 3: Summary of Results

We have found:

\[ x = \frac{3}{5}, \quad y = \frac{1}{\sqrt{5}} \]

### Step 4: Verify the Options

Now we need to check which equation these values satisfy.

1. **First Option:** \( 2x = 1 - y \)
\[ 2 \cdot \frac{3}{5} = \frac{6}{5}, \quad 1 - \frac{1}{\sqrt{5}} \neq \frac{6}{5} \]

2. **Second Option:** \( x^2 = 1 - 2y \)
\[ \left(\frac{3}{5}\right)^2 = \frac{9}{25}, \quad 1 - 2 \cdot \frac{1}{\sqrt{5}} \neq \frac{9}{25} \]

3. **Third Option:** \( x^2 = 1 + 2y \)
\[ \frac{9}{25} \neq 1 + 2 \cdot \frac{1}{\sqrt{5}} \]

4. **Fourth Option:** \( y^2 = 2x - 1 \)
\[ \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5}, \quad 2 \cdot \frac{3}{5} - 1 = \frac{6}{5} - 1 = \frac{1}{5} \]

Thus, the fourth option is correct.