`DeltaDEF` is right angled at E. If `m angle F=60^(@)`, then find the value of `(cot D-2//sqrt(3))`.

Options

A. `1-sqrt(2)`
B. `1//sqrt3`
C. `(2-sqrt(3))//2sqrt(3)`
D. `(2-2sqrt(3))//sqrt(3)`.

Answer

To solve the problem, we need to find the value of \( \cot D - \frac{2}{\sqrt{3}} \) given that triangle \( \Delta DEF \) is right-angled at \( E \) and \( m \angle F = 60^\circ \).

### Step-by-Step Solution:

1. **Identify the Angles**:
Since \( \Delta DEF \) is right-angled at \( E \) and \( m \angle F = 60^\circ \), we can find \( m \angle D \) using the fact that the sum of angles in a triangle is \( 180^\circ \).
\[ m \angle D + m \angle E + m \angle F = 180^\circ \]
Since \( m \angle E = 90^\circ \),
\[ m \angle D + 90^\circ + 60^\circ = 180^\circ \]
\[ m \angle D = 180^\circ - 150^\circ = 30^\circ \]

2. **Find \( \cot D \)**:
We know that \( D = 30^\circ \). The cotangent of an angle is the reciprocal of the tangent.
\[ \cot D = \cot 30^\circ = \frac{1}{\tan 30^\circ} \]
The value of \( \tan 30^\circ \) is \( \frac{1}{\sqrt{3}} \), so:
\[ \cot 30^\circ = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \]

3. **Substitute \( \cot D \) into the Expression**:
Now we substitute \( \cot D \) into the expression \( \cot D - \frac{2}{\sqrt{3}} \):
\[ \cot D - \frac{2}{\sqrt{3}} = \sqrt{3} - \frac{2}{\sqrt{3}} \]

4. **Combine the Terms**:
To combine these terms, we need a common denominator:
\[ \sqrt{3} = \frac{3}{\sqrt{3}} \]
Thus,
\[ \sqrt{3} - \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} - \frac{2}{\sqrt{3}} = \frac{3 - 2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]

5. **Final Answer**:
Therefore, the value of \( \cot D - \frac{2}{\sqrt{3}} \) is:
\[ \frac{1}{\sqrt{3}} \]