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:
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Approach Solution - 1
Given:
\[ \sec^2(\tan^{-1} \alpha) + \csc^2(\cot^{-1} \beta) = 36, \] and \( \alpha + \beta = 8 \). We are asked to find \( \alpha^2 + \beta \).
Step 1: Use the identity for \( \sec^2 \) and \( \csc^2 \)
We know the following identities: \[ \sec^2(\tan^{-1} \alpha) = 1 + \alpha^2 \quad \text{and} \quad \csc^2(\cot^{-1} \beta) = 1 + \beta^2. \] 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 \( \alpha + \beta = 8 \)
Squaring 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 that \( \alpha + \beta = 8 \) 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: \[ \alpha^2 + \beta = 3^2 + 5 = 9 + 5 = 14. \]
Final Answer:
The value of \( \alpha^2 + \beta \) is \( \boxed{14} \).
Approach Solution -2
Given: The following trigonometric equations:
- \( \tan^{-1} \alpha = A \) which implies \( \tan A = \alpha \)
- \( \cot^{-1} \beta = B \) which implies \( \cot B = \beta \)
- \( \sec^2 A + \csc^2 B = 36 \)
- Also, \( \alpha + \beta = 8 \) (Given)
Step 1: Substituting and simplifying the equation
From \( \sec^2 A + \csc^2 B = 36 \), we know: \[ \sec^2 A = 1 + \tan^2 A \quad \text{and} \quad \csc^2 B = 1 + \cot^2 B. \] Thus: \[ 1 + \tan^2 A + 1 + \cot^2 B = 36 \] Simplifying: \[ \tan^2 A + \cot^2 B = 34 \] Therefore, we get the equation: \[ \alpha^2 + \beta^2 = 34 \quad \text{(since \( \tan A = \alpha \) and \( \cot B = \beta \))}. \]
Step 2: Using the sum of \( \alpha \) and \( \beta \)
It is given that: \[ \alpha + \beta = 8. \] So: \[ (\alpha + \beta)^2 = 34 + 2\alpha\beta. \] Substituting \( \alpha + \beta = 8 \) into this equation: \[ 64 = 34 + 2\alpha\beta \] \[ 2\alpha\beta = 30 \] \[ \alpha\beta = 15. \]
Step 3: Solving the quadratic equation
Now, \( \alpha \) and \( \beta \) are the roots of the quadratic equation: \[ x^2 - 8x + 15 = 0. \] We can solve this quadratic equation using the factorization method: \[ (x - 3)(x - 5) = 0. \] Thus, the solutions are: \[ x = 3 \quad \text{or} \quad x = 5. \] So, \( \alpha = 3 \) and \( \beta = 5 \), with \( \alpha < \beta \).
Step 4: Final calculation of \( \alpha^2 + \beta^2 \)
Now, we can calculate: \[ \alpha^2 + \beta^2 = 3^2 + 5^2 = 9 + 25 = 14. \]
Final Answer:
\[ \boxed{\alpha^2 + \beta^2 = 14} \]
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