If $ a \cot \theta + b \csc \theta = p $ and $ b \cot \theta + a \csc \theta = q $, then find the value of $ p^2 - q^2 $.

Question

Accepted Answer

**Solution:**

1. Given:
   $$
   p = a \cot \theta + b \csc \theta
   $$
   $$
   q = b \cot \theta + a \csc \theta
   $$

2. Square both expressions:
   $$
   p^2 = (a \cot \theta + b \csc \theta)^2 = a^2 \cot^2 \theta + b^2 \csc^2 \theta + 2ab \cot \theta \csc \theta
   $$
   $$
   q^2 = (b \cot \theta + a \csc \theta)^2 = b^2 \cot^2 \theta + a^2 \csc^2 \theta + 2ab \cot \theta \csc \theta
   $$

3. Subtract $ q^2 $ from $ p^2 $:
   $$
   p^2 - q^2 = (a^2 \cot^2 \theta + b^2 \csc^2 \theta + 2ab \cot \theta \csc \theta) - (b^2 \cot^2 \theta + a^2 \csc^2 \theta + 2ab \cot \theta \csc \theta)
   $$

4. Simplify by canceling common terms:
   $$
   p^2 - q^2 = a^2 \cot^2 \theta - b^2 \cot^2 \theta + b^2 \csc^2 \theta - a^2 \csc^2 \theta
   $$

5. Group terms:
   $$
   p^2 - q^2 = a^2 (\cot^2 \theta - \csc^2 \theta) + b^2 (\csc^2 \theta - \cot^2 \theta)
   $$

6. Use the Pythagorean identity:
   $$
   1 + \cot^2 \theta = \csc^2 \theta \implies \cot^2 \theta = \csc^2 \theta - 1
   $$
   $$
   \csc^2 \theta - \cot^2 \theta = 1
   $$

7. Substitute these into the expression:
   $$
   p^2 - q^2 = a^2 (-1) + b^2 (1) = -a^2 + b^2 = b^2 - a^2
   $$

**Answer:** $b^2 - a^2$

Answer

Answer

Answer