`DeltaDEF` is right angled at E .If `mangleD=60^(@)` ,then find the value of `(cotF+(1)/(sqrt(3) )

Options

A. `(3sqrt(3))/(2)`
B. `(4)/(sqrt(3))`
C. `((2+2sqrt(3)))/(sqrt(3))`
D. `(7)/(2sqrt(3))`

Answer

To solve the problem, we need to find the value of \( \cot F + \frac{1}{\sqrt{3}} \) given that triangle DEF is a right triangle at E, and \( \angle D = 60^\circ \).

### Step-by-Step Solution:

1. **Identify the Angles in Triangle DEF**:
Since triangle DEF is a right triangle at E, we know:
\[ \angle D + \angle E + \angle F = 180^\circ \]
Given that \( \angle E = 90^\circ \) and \( \angle D = 60^\circ \), we can substitute these values into the equation:
\[ 60^\circ + 90^\circ + \angle F = 180^\circ \]

2. **Calculate Angle F**:
Rearranging the equation gives:
\[ \angle F = 180^\circ - 60^\circ - 90^\circ \]
Simplifying this, we find:
\[ \angle F = 30^\circ \]

3. **Find the Value of \( \cot F \)**:
The cotangent of an angle is defined as the reciprocal of the tangent:
\[ \cot F = \frac{1}{\tan F} \]
We know that:
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Therefore:
\[ \cot 30^\circ = \sqrt{3} \]

4. **Calculate \( \cot F + \frac{1}{\sqrt{3}} \)**:
Now we can substitute \( \cot F \) into the expression:
\[ \cot F + \frac{1}{\sqrt{3}} = \sqrt{3} + \frac{1}{\sqrt{3}} \]
To combine these terms, we can express \( \sqrt{3} \) as \( \frac{3}{\sqrt{3}} \):
\[ \cot F + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{3 + 1}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]

5. **Final Answer**:
Thus, the value of \( \cot F + \frac{1}{\sqrt{3}} \) is:
\[ \frac{4}{\sqrt{3}} \]