To solve the equation \( \tan^3 \theta - \tan \theta - 1 = A \sec^2 \theta + B \tan \theta \), we first need to manipulate the left-hand side to match the right-hand side.
Step 1: Express in terms of \( \tan \theta \)
- Let \( x = \tan \theta \).
- The equation becomes \( x^3 - x - 1 = A(1 + x^2) + Bx \) since \( \sec^2 \theta = 1 + \tan^2 \theta = 1 + x^2 \).
Step 2: Expand the right-hand side
- Expanding gives:
\[
A(1 + x^2) + Bx = A + Ax^2 + Bx
\]
- Therefore, the equation can be rewritten as:
\[
x^3 - x - 1 = Ax^2 + Bx + A
\]
Step 3: Rearranging the equation
- Rearranging yields:
\[
x^3 - Ax^2 - (B + 1)x - (A + 1) = 0
\]
Step 4: Coefficients Comparison
- For the equation to hold for all \( x \), coefficients of \( x^3, x^2, x^1, \) and constant terms must be equal.
- Coefficient of \( x^3 \): \( 1 \)
- Coefficient of \( x^2 \): \( -A = 0 \) → \( A = 0 \)
- Coefficient of \( x \): \( -(B + 1) = -1 \) → \( B + 1 = 1 \) → \( B = 0 \)
- Constant: \( -(A + 1) = -1 \) → \( A + 1 = 1 \)
Final Step: Calculate \( A + B \)
- Since \( A = 0 \) and \( B = 0 \):
\[
A + B = 0 + 0 = 0
\]
Thus, the final answer is \( A + B = 0 \).