If `btantheta=a`, the value of `(asintheta-bcostheta)/(asintheta+bcostheta)`

Options

A. `(a-b)/(a^2+b^2)`
B. `(a+b)/(a^2+b^2)`
C. `(a^2+b^2)/(a^2-b^2)`
D. `(a^2-b^2)/(a^2+b^2)`

Answer

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

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

We need to find the value of:
\[ \frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} \]

### Step 1: Express \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\)

From the definition of tangent, we know:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Thus, we can express \(\sin \theta\) in terms of \(\tan \theta\) and \(\cos \theta\):
\[ \sin \theta = \tan \theta \cdot \cos \theta \]

### Step 2: Substitute \(\tan \theta\)

Substituting \(\tan \theta = \frac{b}{a}\) into the equation for \(\sin \theta\):
\[ \sin \theta = \frac{b}{a} \cdot \cos \theta \]

### Step 3: Substitute \(\sin \theta\) into the original expression

Now we substitute \(\sin \theta\) into the expression:
\[ \frac{a \left(\frac{b}{a} \cos \theta\right) - b \cos \theta}{a \left(\frac{b}{a} \cos \theta\right) + b \cos \theta} \]

### Step 4: Simplify the expression

This simplifies to:
\[ \frac{b \cos \theta - b \cos \theta}{b \cos \theta + b \cos \theta} \]
This can be rewritten as:
\[ \frac{b \cos \theta (1 - 1)}{b \cos \theta (1 + 1)} = \frac{0}{2b \cos \theta} = 0 \]

### Step 5: Final Expression

Thus, the final value of the expression is:
\[ \frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} = 0 \]

### Summary

The value of \(\frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta}\) is \(0\).