If `x=sin(2tan^(-1)2sqrt3)and y=sin((1)/(2)tan^(-1).(12)/(5))`, then

Options

A. `x=1-y`
B. `x^(2)=1-2y`
C. `x^(2)=1+y`
D. `y^(2)=(x)/(sqrt3)`

Answer

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

\[ x = \sin(2 \tan^{-1}(2\sqrt{3})) \]
\[ y = \sin\left(\frac{1}{2} \tan^{-1}\left(\frac{12}{5}\right)\right) \]

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

Let \( \theta = \tan^{-1}(2\sqrt{3}) \). Then, we have:

\[ \tan \theta = 2\sqrt{3} \]

Using the double angle identity for sine, we have:

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

Now, we need to find \( \sin \theta \) and \( \cos \theta \).

### Step 2: Find \( \sin \theta \) and \( \cos \theta \)

From \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{2\sqrt{3}}{1} \), we can form a right triangle where:

- Opposite side = \( 2\sqrt{3} \)
- Adjacent side = \( 1 \)

Using the Pythagorean theorem to find the hypotenuse \( h \):

\[ h = \sqrt{(2\sqrt{3})^2 + 1^2} = \sqrt{12 + 1} = \sqrt{13} \]

Thus, we can find \( \sin \theta \) and \( \cos \theta \):

\[ \sin \theta = \frac{2\sqrt{3}}{\sqrt{13}}, \quad \cos \theta = \frac{1}{\sqrt{13}} \]

### Step 3: Substitute into the expression for \( x \)

Now substituting back into the expression for \( x \):

\[ x = 2 \sin \theta \cos \theta = 2 \left(\frac{2\sqrt{3}}{\sqrt{13}}\right) \left(\frac{1}{\sqrt{13}}\right) = \frac{4\sqrt{3}}{13} \]

### Step 4: Calculate \( y \)

Let \( \alpha = \tan^{-1}\left(\frac{12}{5}\right) \). Then, we have:

\[ \tan \alpha = \frac{12}{5} \]

Using the half angle identity for sine, we have:

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

### Step 5: Find \( \cos \alpha \)

From \( \tan \alpha = \frac{12}{5} \), we can form a right triangle where:

- Opposite side = \( 12 \)
- Adjacent side = \( 5 \)

Using the Pythagorean theorem to find the hypotenuse \( h \):

\[ h = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

Thus, we can find \( \cos \alpha \):

\[ \cos \alpha = \frac{5}{13} \]

### Step 6: Substitute into the expression for \( y \)

Now substituting back into the expression for \( y \):

\[ y = \sqrt{\frac{1 - \frac{5}{13}}{2}} = \sqrt{\frac{\frac{8}{13}}{2}} = \sqrt{\frac{8}{26}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}} \]

### Step 7: Relate \( x \) and \( y \)

Now we have:

\[ x = \frac{4\sqrt{3}}{13}, \quad y = \frac{2}{\sqrt{13}} \]

To find the relationship, we can square \( y \):

\[ y^2 = \left(\frac{2}{\sqrt{13}}\right)^2 = \frac{4}{13} \]

Now, we can express \( x \) in terms of \( y^2 \):

\[ x = \frac{4\sqrt{3}}{13} \]

Thus, we can write:

\[ y^2 = \frac{x}{\sqrt{3}} \]

### Final Relation

The final relation between \( x \) and \( y \) is:

\[ y^2 = \frac{x}{\sqrt{3}} \]