If tan `theta = a/b` then the value of `(b sin theta - a cos theta)/(b sin theta + a cos theta)` is

Options

A. 1
B. `(a^2-b^2)/(a^2+b^2)`
C. `(b^2-a^2)/(b^2+a^2)`
D. 0

Answer

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

1. We know that \( \tan \theta = \frac{a}{b} \).

From the definition of tangent, we can express this in terms of sine and cosine:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]

Thus, we have:

\[ \frac{\sin \theta}{\cos \theta} = \frac{a}{b} \]

From this, we can express \( \sin \theta \) and \( \cos \theta \):

\[ \sin \theta = \frac{a}{\sqrt{a^2 + b^2}} \quad \text{and} \quad \cos \theta = \frac{b}{\sqrt{a^2 + b^2}} \]

Now, we need to find the value of:

\[ \frac{b \sin \theta - a \cos \theta}{b \sin \theta + a \cos \theta} \]

Substituting the expressions for \( \sin \theta \) and \( \cos \theta \):

\[ b \sin \theta = b \cdot \frac{a}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}} \]

\[ a \cos \theta = a \cdot \frac{b}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}} \]

Now, substituting these into our expression:

1. **Numerator**:
\[ b \sin \theta - a \cos \theta = \frac{ab}{\sqrt{a^2 + b^2}} - \frac{ab}{\sqrt{a^2 + b^2}} = 0 \]

2. **Denominator**:
\[ b \sin \theta + a \cos \theta = \frac{ab}{\sqrt{a^2 + b^2}} + \frac{ab}{\sqrt{a^2 + b^2}} = \frac{2ab}{\sqrt{a^2 + b^2}} \]

Now substituting back into our expression:

\[ \frac{0}{\frac{2ab}{\sqrt{a^2 + b^2}}} = 0 \]

Thus, the final value is:

\[ \boxed{0} \]