The value of `tan7 1/2^@` equal to

Options

A. `(sqrt(3)-sqrt(2))/(sqrt(2)+1)`
B. `(sqrt(3)-sqrt(2))/(sqrt(2)-1)`
C. `(sqrt(3)+sqrt(2))/(sqrt(2)+1)`
D. `(sqrt(3)+sqrt(2))/(sqrt(2)-1)`

Answer

To find the value of \( \tan 7.5^\circ \), we can use the half-angle formula for tangent, which states:

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

In this case, we will use \( x = 15^\circ \) since \( 7.5^\circ = \frac{15^\circ}{2} \).

### Step 1: Calculate \( \cos 15^\circ \) and \( \sin 15^\circ \)

Using the angle subtraction formulas:

\[ \sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \]

Substituting the values:

\[ \sin 15^\circ = \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

Now for \( \cos 15^\circ \):

\[ \cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \]

Substituting the values:

\[ \cos 15^\circ = \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

### Step 2: Substitute into the Half-Angle Formula

Now we can substitute \( \sin 15^\circ \) and \( \cos 15^\circ \) into the half-angle formula:

\[ \tan 7.5^\circ = \frac{1 - \cos 15^\circ}{\sin 15^\circ} \]

Substituting the values we found:

\[ \tan 7.5^\circ = \frac{1 - \frac{\sqrt{3} + 1}{2\sqrt{2}}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} \]

### Step 3: Simplify the Expression

First, simplify the numerator:

\[ 1 - \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{2\sqrt{2}}{2\sqrt{2}} - \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{2\sqrt{2} - (\sqrt{3} + 1)}{2\sqrt{2}} = \frac{2\sqrt{2} - \sqrt{3} - 1}{2\sqrt{2}} \]

Now substitute this back into the tangent expression:

\[ \tan 7.5^\circ = \frac{\frac{2\sqrt{2} - \sqrt{3} - 1}{2\sqrt{2}}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} \]

This simplifies to:

\[ \tan 7.5^\circ = \frac{2\sqrt{2} - \sqrt{3} - 1}{\sqrt{3} - 1} \]

### Step 4: Rationalize the Denominator

To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} + 1 \):

\[ \tan 7.5^\circ = \frac{(2\sqrt{2} - \sqrt{3} - 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(2\sqrt{2} - \sqrt{3} - 1)(\sqrt{3} + 1)}{3 - 1} = \frac{(2\sqrt{2} - \sqrt{3} - 1)(\sqrt{3} + 1)}{2} \]

### Final Expression

After simplifying the numerator, we can find the final value of \( \tan 7.5^\circ \).

### Summary of Steps

1. Use angle subtraction formulas to find \( \sin 15^\circ \) and \( \cos 15^\circ \).
2. Substitute these values into the half-angle formula for tangent.
3. Simplify the expression.
4. Rationalize the denominator if necessary.