If $tan~ heta=\frac{p}{q}$ then $\frac{p~sin~ heta-q~cos~ heta}{p~sin~ heta+q~cos~ heta}$ is equal to:

Question

If $tan~	heta=\frac{p}{q}$ then $\frac{p~sin~	heta-q~cos~	heta}{p~sin~	heta+q~cos~	heta}$ is equal to:

Accepted Answer

To solve the given problem, we need to evaluate the expression $\frac{p \sin \theta - q \cos \theta}{p \sin \theta + q \cos \theta}$ given that $\tan \theta = \frac{p}{q}$.

First, let's express $\sin \theta$ and $\cos \theta$ in terms of $p$ and $q$. Since $\tan \theta = \frac{p}{q}$, we have:

$$
\sin \theta = \frac{p}{\sqrt{p^2 + q^2}}
$$
$$
\cos \theta = \frac{q}{\sqrt{p^2 + q^2}}
$$

Now, substitute these into the expression:

$$
\frac{p \sin \theta - q \cos \theta}{p \sin \theta + q \cos \theta} = \frac{p \left(\frac{p}{\sqrt{p^2 + q^2}}\right) - q \left(\frac{q}{\sqrt{p^2 + q^2}}\right)}{p \left(\frac{p}{\sqrt{p^2 + q^2}}\right) + q \left(\frac{q}{\sqrt{p^2 + q^2}}\right)}
$$

Simplify the expression:

$$
= \frac{\frac{p^2}{\sqrt{p^2 + q^2}} - \frac{q^2}{\sqrt{p^2 + q^2}}}{\frac{p^2}{\sqrt{p^2 + q^2}} + \frac{q^2}{\sqrt{p^2 + q^2}}}
$$

$$
= \frac{p^2 - q^2}{p^2 + q^2}
$$

Thus, the expression simplifies to $\frac{p^2 - q^2}{p^2 + q^2}$, which matches Option 2.

Now, let's explain why the other options are incorrect:

1. **Option 1: $\frac{p^2 - q^2}{p + q^2}$**

This option is incorrect because the denominator should be $p^2 + q^2$, not $p + q^2$. The expression derived from the trigonometric identities clearly shows that both terms in the denominator are squared.

2. **Option 3: $\frac{p - q^2}{p^2 + q^2}$**

This option is incorrect because the numerator should be $p^2 - q^2$, not $p - q^2$. The correct simplification involves squaring both $p$ and $q$ in the numerator.

3. **Option 4: $\frac{p^2}{q^2}$**

This option is incorrect because it does not account for the subtraction of $q^2$ from $p^2$ in the numerator, nor does it correctly represent the denominator as $p^2 + q^2$.

In conclusion, the correct answer is Option 2: $\frac{p^2 - q^2}{p^2 + q^2}$, as it accurately reflects the simplification of the given expression using the trigonometric identities for $\sin \theta$ and $\cos \theta$.