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

We are given: \[ \sec^2(\tan^{-1} \alpha) + \csc^2(\cot^{-1} \beta) = 36, \] and \( \alpha + \beta = 8 \), and we need to find \( \alpha^2 + \beta \).
Step 1: Use the identity for \( \sec^2 \) and \( \csc^2 \). We know that: \[ \sec^2(\tan^{-1} \alpha) = 1 + \alpha^2 \quad {and} \quad \csc^2(\cot^{-1} \beta) = 1 + \beta^2. \] Thus, the given equation becomes: \[ 1 + \alpha^2 + 1 + \beta^2 = 36. \] Simplifying: \[ \alpha^2 + \beta^2 + 2 = 36 \quad \Rightarrow \quad \alpha^2 + \beta^2 = 34. \] 
Step 2: Use the given condition \( \alpha + \beta = 8 \). Square both sides of \( \alpha + \beta = 8 \): \[ (\alpha + \beta)^2 = 64. \] Expanding: \[ \alpha^2 + 2\alpha \beta + \beta^2 = 64. \] We already know that \( \alpha^2 + \beta^2 = 34 \), so substitute this into the equation: \[ 34 + 2\alpha \beta = 64 \quad \Rightarrow \quad 2\alpha \beta = 30 \quad \Rightarrow \quad \alpha \beta = 15. \] 
Step 3: Solve for \( \alpha^2 + \beta \). We know \( \alpha + \beta = 8 \) and \( \alpha \beta = 15 \). We can now use the quadratic equation whose roots are \( \alpha \) and \( \beta \): \[ t^2 - (\alpha + \beta)t + \alpha \beta = 0 \quad \Rightarrow \quad t^2 - 8t + 15 = 0. \] The solutions for \( t \) are given by: \[ t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(15)}}{2(1)} = \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm \sqrt{4}}{2} = \frac{8 \pm 2}{2}. \] Thus, \( t = 5 \) or \( t = 3 \). So, \( \alpha = 3 \) and \( \beta = 5 \) (since \( \alpha \leq \beta \)). 
Step 4: Compute \( \alpha^2 + \beta \). Now that we know \( \alpha = 3 \) and \( \beta = 5 \), we compute: \[ \alpha^2 + \beta = 3^2 + 5 = 9 + 5 = 14. \] Thus, the value of \( \alpha^2 + \beta \) is \( \boxed{14} \).