Question

If for some \( \alpha, \beta \); \( \alpha \leq \beta \), \( \alpha + \beta = 8 \) and \[ \sec^2(\tan^{-1} \alpha) + \csc^2(\cot^{-1} \beta) = 36, \] then \( \alpha^2 + \beta \) is:

Hint:

When solving problems involving inverse trigonometric functions and their squares, use the identities \( \sec^2(\tan^{-1} x) = 1 + x^2 \) and \( \csc^2(\cot^{-1} x) = 1 + x^2 \). Then use the sum of squares and sum of products to solve for the required values.

Answer 1

Given:

\[ \sec^2(\tan^{-1} \alpha) + \csc^2(\cot^{-1} \beta) = 36, \] and \( \alpha + \beta = 8 \). Find \( \alpha^2 + \beta \).

Step 1: Apply trigonometric identities

Using the identities \( \sec^2(\tan^{-1} \alpha) = 1 + \alpha^2 \) and \( \csc^2(\cot^{-1} \beta) = 1 + \beta^2 \), the given equation simplifies to: \[ 1 + \alpha^2 + 1 + \beta^2 = 36 \] which reduces to \[ \alpha^2 + \beta^2 = 34. \]

Step 2: Utilize the sum condition

Squaring the equation \( \alpha + \beta = 8 \) gives \( (\alpha + \beta)^2 = 64 \), which expands to \[ \alpha^2 + 2\alpha \beta + \beta^2 = 64. \] Substituting \( \alpha^2 + \beta^2 = 34 \): \[ 34 + 2\alpha \beta = 64 \quad \Rightarrow \quad 2\alpha \beta = 30 \quad \Rightarrow \quad \alpha \beta = 15. \]

Step 3: Determine \( \alpha \) and \( \beta \)

With \( \alpha + \beta = 8 \) and \( \alpha \beta = 15 \), \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( t^2 - 8t + 15 = 0 \). Solving this equation yields \( t = 3 \) or \( t = 5 \). Assuming \( \alpha \leq \beta \), we have \( \alpha = 3 \) and \( \beta = 5 \).

Step 4: Calculate \( \alpha^2 + \beta \)

Substitute the values of \( \alpha \) and \( \beta \) into the expression: \[ \alpha^2 + \beta = 3^2 + 5 = 9 + 5 = 14. \]

Final Answer:

The value of \( \alpha^2 + \beta \) is \( \boxed{14} \).