If for some α,β; α≤β, α+β=8 and sec2(tan−1α)+csc2(cot−1β)=36, then α2+β is
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Step-by-Step Solution
Step 1: Simplify the trigonometric expression
We use the identities sec2x=1+tan2x and csc2x=1+cot2x. Let x=tan−1α and y=cot−1β. Then tanx=α and coty=β. Substituting these into the given equation simplifies the expression.
Step 2: Substitute and simplify
After applying the identities, we substitute tan(tan−1α)=α and cot(cot−1β)=β. This leads to a simplified algebraic equation involving α2 and β2.
Step 3: Use the sum of roots identity
We are given α+β=8 and we found α2+β2=34. We can use the algebraic identity for the square of a sum to find the product αβ.
Step 4: Calculate the product αβ
Substitute the known values into the identity. This allows us to solve for 2αβ and then for αβ.
Step 5: Form a quadratic equation for α and β
We can form a quadratic equation whose roots are α and β using the sum and product of the roots. The general form is x2−(sum of roots)x+(product of roots)=0.
Step 6: Solve the quadratic equation
Factor the quadratic equation to find the possible values for x. These values represent α and β.
Step 7: Determine α and β and calculate α2+β
Since it is given that α≤β, we assign α=3 and β=5. Finally, we calculate the required expression α2+β.