If cosec ` theta = sqrt2 `, find the value of :
` (1)/(tan A ) +( sin A)/(1+ cos A) `

Answer

To solve the problem, we start with the given information that \( \csc \theta = \sqrt{2} \).

### Step 1: Understand the relationship of cosecant
Cosecant is defined as the reciprocal of sine:
\[ \csc \theta = \frac{1}{\sin \theta} \]
Thus, if \( \csc \theta = \sqrt{2} \), we can find \( \sin \theta \):
\[ \sin \theta = \frac{1}{\csc \theta} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Step 2: Find cosine using the Pythagorean identity
Using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
we can find \( \cos \theta \):
\[ \left(\frac{\sqrt{2}}{2}\right)^2 + \cos^2 \theta = 1 \]
\[ \frac{2}{4} + \cos^2 \theta = 1 \]
\[ \frac{1}{2} + \cos^2 \theta = 1 \]
\[ \cos^2 \theta = 1 - \frac{1}{2} = \frac{1}{2} \]
\[ \cos \theta = \pm \frac{\sqrt{2}}{2} \]

### Step 3: Find tangent
Tangent is defined as:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Using \( \sin \theta = \frac{\sqrt{2}}{2} \) and \( \cos \theta = \frac{\sqrt{2}}{2} \):
\[ \tan \theta = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 \]

### Step 4: Substitute values into the expression
Now we need to evaluate:
\[ \frac{1}{\tan A} + \frac{\sin A}{1 + \cos A} \]
Substituting the values we found:
\[ \frac{1}{1} + \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} \]
This simplifies to:
\[ 1 + \frac{\frac{\sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}} = 1 + \frac{\sqrt{2}}{2 + \sqrt{2}} \]

### Step 5: Simplify the second term
To simplify \( \frac{\sqrt{2}}{2 + \sqrt{2}} \), we can rationalize the denominator:
\[ \frac{\sqrt{2}}{2 + \sqrt{2}} \cdot \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{\sqrt{2}(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} \]
The denominator simplifies to:
\[ (2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2 \]
Thus, we have:
\[ \frac{\sqrt{2}(2 - \sqrt{2})}{2} = \frac{2\sqrt{2} - 2}{2} = \sqrt{2} - 1 \]

### Step 6: Combine the terms
Now we can combine the terms:
\[ 1 + \left(\sqrt{2} - 1\right) = \sqrt{2} \]

### Final Answer
Thus, the final value is:
\[ \sqrt{2} \]