`DeltaABC` is right angled at B. If `mangleA = 45^@` , then find the value of `("cosecC "+ 1//sqrt3)`.

Options

A. `(3sqrt2+1)//3`
B. `(2+sqrt3)//2`
C. `5//sqrt3`
D. `(sqrt6 +1)//sqrt3`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Understand the triangle properties
Given that triangle ABC is a right triangle with the right angle at B, and angle A is 45 degrees, we can conclude that angle C must also be 45 degrees because the sum of angles in a triangle is 180 degrees.

### Step 2: Calculate angle C
Since angle A = 45 degrees and angle B = 90 degrees, we can find angle C as follows:
\[ \text{Angle C} = 180^\circ - \text{Angle A} - \text{Angle B} = 180^\circ - 45^\circ - 90^\circ = 45^\circ \]

### Step 3: Find cosec C
Now that we know angle C is 45 degrees, we can find cosec C:
\[ \text{cosec C} = \text{cosec}(45^\circ) \]
The cosecant of 45 degrees is:
\[ \text{cosec}(45^\circ) = \frac{1}{\sin(45^\circ)} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]

### Step 4: Add 1/√3 to cosec C
Next, we need to calculate:
\[ \text{cosec C} + \frac{1}{\sqrt{3}} = \sqrt{2} + \frac{1}{\sqrt{3}} \]

### Step 5: Find a common denominator
To add these two fractions, we need a common denominator. The least common multiple of √2 and √3 is √6. Therefore, we rewrite each term:
\[ \sqrt{2} = \frac{\sqrt{2} \cdot \sqrt{3}}{\sqrt{6}} = \frac{\sqrt{6}}{\sqrt{6}} \]
\[ \frac{1}{\sqrt{3}} = \frac{1 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{3}} = \frac{\sqrt{2}}{\sqrt{6}} \]

### Step 6: Combine the fractions
Now we can combine the two fractions:
\[ \sqrt{2} + \frac{1}{\sqrt{3}} = \frac{\sqrt{6}}{\sqrt{6}} + \frac{\sqrt{2}}{\sqrt{6}} = \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6}} \]

### Final Answer
Thus, the final value of \(\text{cosec C} + \frac{1}{\sqrt{3}}\) is:
\[ \sqrt{2} + \frac{1}{\sqrt{3}} = \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6}} \]