(i) If `tantheta=(a)/(b)`,then find the value of `(2sintheta-3costheta)/(2sintheta+3costheta)`

Answer

To solve the problem, we start with the given equation:

\[ \tan \theta = \frac{a}{b} \]

We need to find the value of:

\[ \frac{2 \sin \theta - 3 \cos \theta}{2 \sin \theta + 3 \cos \theta} \]

### Step 1: Rewrite the expression using sine and cosine
We can rewrite the expression by dividing both the numerator and the denominator by \(\cos \theta\):

\[ \frac{2 \sin \theta - 3 \cos \theta}{2 \sin \theta + 3 \cos \theta} = \frac{\frac{2 \sin \theta}{\cos \theta} - \frac{3 \cos \theta}{\cos \theta}}{\frac{2 \sin \theta}{\cos \theta} + \frac{3 \cos \theta}{\cos \theta}} \]

### Step 2: Simplify the fractions
This simplifies to:

\[ \frac{2 \tan \theta - 3}{2 \tan \theta + 3} \]

### Step 3: Substitute \(\tan \theta\)
Now, we substitute \(\tan \theta\) with \(\frac{a}{b}\):

\[ \frac{2 \left(\frac{a}{b}\right) - 3}{2 \left(\frac{a}{b}\right) + 3} \]

### Step 4: Clear the fraction
To eliminate the fraction, we multiply the numerator and the denominator by \(b\):

\[ \frac{2a - 3b}{2a + 3b} \]

### Final Result
Thus, the value of \(\frac{2 \sin \theta - 3 \cos \theta}{2 \sin \theta + 3 \cos \theta}\) is:

\[ \frac{2a - 3b}{2a + 3b} \]

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