If `tan theta=a/b` and `0 to theta lt pi`, then the value of `(sqrt(a^(2)+b^(2))/(|a|) sin theta)` is ………..(Assume, `b ne 0, a ne 0, a,b in R`)

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{\sqrt{a^2 + b^2}}{|a| \sin \theta} \]

given that \(\tan \theta = \frac{a}{b}\) and \(0 < \theta < \pi\).

### Step-by-step Solution:

1. **Start with the given relationship**:
\[ \tan \theta = \frac{a}{b} \]
This implies that:
\[ \frac{b}{a} = \frac{1}{\tan \theta} = \cot \theta \]

2. **Rewrite the expression**:
We need to evaluate:
\[ \frac{\sqrt{a^2 + b^2}}{|a| \sin \theta} \]

3. **Express \(b\) in terms of \(a\) and \(\tan \theta\)**:
From \(\tan \theta = \frac{a}{b}\), we can express \(b\) as:
\[ b = \frac{a}{\tan \theta} \]

4. **Substitute \(b\) into the expression**:
Substitute \(b\) into the expression \(a^2 + b^2\):
\[ a^2 + b^2 = a^2 + \left(\frac{a}{\tan \theta}\right)^2 = a^2 + \frac{a^2}{\tan^2 \theta} = a^2 \left(1 + \frac{1}{\tan^2 \theta}\right) \]

5. **Use the identity for \(\tan\)**:
Recall that:
\[ 1 + \frac{1}{\tan^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta} \]
Thus:
\[ a^2 + b^2 = a^2 \cdot \frac{1}{\sin^2 \theta} = \frac{a^2}{\sin^2 \theta} \]

6. **Take the square root**:
Now, we take the square root:
\[ \sqrt{a^2 + b^2} = \sqrt{\frac{a^2}{\sin^2 \theta}} = \frac{|a|}{|\sin \theta|} \]

7. **Substitute back into the expression**:
Now substituting back into the original expression:
\[ \frac{\sqrt{a^2 + b^2}}{|a| \sin \theta} = \frac{\frac{|a|}{|\sin \theta|}}{|a| \sin \theta} = \frac{1}{|\sin \theta| \sin \theta} \]

8. **Simplify the expression**:
Since \(|\sin \theta| = \sin \theta\) for \(0 < \theta < \pi\):
\[ \frac{1}{\sin^2 \theta} \]

9. **Final Result**:
Therefore, the value of the expression is:
\[ 1 \]