If `p sintheta+qcostheta=aandpcostheta-qsintheta=b,` show that : `(p+a)/(q+b)+(q-b)/(p-a)=0`.

Answer

To prove the equation \(\frac{p + a}{q + b} + \frac{q - b}{p - a} = 0\) given the equations \(p \sin \theta + q \cos \theta = a\) and \(p \cos \theta - q \sin \theta = b\), we can follow these steps:

### Step 1: Square both equations
We start with the two equations:
1. \(p \sin \theta + q \cos \theta = a\)
2. \(p \cos \theta - q \sin \theta = b\)

Now, we square both equations:

\[ (p \sin \theta + q \cos \theta)^2 = a^2 \]
\[ (p \cos \theta - q \sin \theta)^2 = b^2 \]

### Step 2: Expand the squares
Expanding both squared equations gives us:

1. \(p^2 \sin^2 \theta + 2pq \sin \theta \cos \theta + q^2 \cos^2 \theta = a^2\)
2. \(p^2 \cos^2 \theta - 2pq \sin \theta \cos \theta + q^2 \sin^2 \theta = b^2\)

### Step 3: Combine the equations
Now, we add both equations:

\[ (p^2 \sin^2 \theta + q^2 \cos^2 \theta + 2pq \sin \theta \cos \theta) + (p^2 \cos^2 \theta + q^2 \sin^2 \theta - 2pq \sin \theta \cos \theta) = a^2 + b^2 \]

### Step 4: Simplify the left side
Notice that \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ p^2 (\sin^2 \theta + \cos^2 \theta) + q^2 (\sin^2 \theta + \cos^2 \theta) = a^2 + b^2 \]
\[ p^2 + q^2 = a^2 + b^2 \]

### Step 5: Rearranging the terms
From \(p^2 + q^2 = a^2 + b^2\), we can rearrange it as:

\[ p^2 - a^2 = b^2 - q^2 \]

### Step 6: Factor the differences of squares
This can be factored as:

\[ (p - a)(p + a) = (b - q)(b + q) \]

### Step 7: Rearranging to prove the expression
From the above equation, we can express:

\[ \frac{p + a}{q + b} = -\frac{q - b}{p - a} \]

### Step 8: Combine the fractions
Now, we can write:

\[ \frac{p + a}{q + b} + \frac{q - b}{p - a} = 0 \]

Thus, we have shown that:

\[ \frac{p + a}{q + b} + \frac{q - b}{p - a} = 0 \]

### Conclusion
Hence, the required expression is proved.

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