`DeltaXYZ` is right angled at Y. If `mangleX = 60^@`, then find the value of `(secz + 2/ sqrt3)`.

Options

A. `4// sqrt3`
B. `(sqrt2+2)//2sqrt2`
C. `7//2sqrt3`
D. `4//2sqrt3`

Answer

To solve the problem, we need to find the value of \( \sec Z + \frac{2}{\sqrt{3}} \) given that triangle \( \Delta XYZ \) is right-angled at \( Y \) and \( \angle X = 60^\circ \).

### Step-by-Step Solution:

1. **Identify the Angles in Triangle XYZ**:
Since triangle \( XYZ \) is right-angled at \( Y \) and \( \angle X = 60^\circ \), we can find \( \angle Z \).
\[ \angle Y = 90^\circ, \quad \angle X = 60^\circ \]
The sum of angles in a triangle is \( 180^\circ \):
\[ \angle X + \angle Y + \angle Z = 180^\circ \]
Substituting the known angles:
\[ 60^\circ + 90^\circ + \angle Z = 180^\circ \]
\[ \angle Z = 180^\circ - 150^\circ = 30^\circ \]

2. **Calculate \( \sec Z \)**:
The secant function is defined as the reciprocal of the cosine function:
\[ \sec Z = \frac{1}{\cos Z} \]
For \( \angle Z = 30^\circ \):
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]
Therefore:
\[ \sec Z = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \]

3. **Add \( \sec Z \) and \( \frac{2}{\sqrt{3}} \)**:
Now we need to find \( \sec Z + \frac{2}{\sqrt{3}} \):
\[ \sec Z + \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{2 + 2}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]

### Final Answer:
Thus, the value of \( \sec Z + \frac{2}{\sqrt{3}} \) is \( \frac{4}{\sqrt{3}} \).