If for some alpha, beta such that alpha leq beta, alpha + beta = 8 and sec^2(tan^-1alpha) + csc^2(cot^-1beta) = 36, then alpha^2 + beta is equal to ________.

Numerical Answer Type:
Enter a numerical value Answer: 14 +4 marks

Solution & Explanation

### Related Formula Apply the fundamental trigonometric identity properties directly linking reciprocal functions: sec^2(theta) = 1 + tan^2(theta) csc^2(phi) = 1 + cot^2(phi) ### Core Logic Simplify the given trigonometric equation using the identity formulas: sec^2(tan^-1alpha) = 1 + tan^2(tan^-1alpha) = 1 + alpha^2 csc^2(cot^-1beta) = 1 + cot^2(cot^-1beta) = 1 + beta^2 Substitute these simplified expressions back into the target relation equation: (1 + alpha^2) + (1 + beta^2) = 36 implies alpha^2 + beta^2 = 34 ### Step 1: Set up a Quadratic Equation for the roots We are given the linear \sum alpha + beta = 8. Use the algebraic identity for squares to find the product: (alpha + beta)^2 = alpha^2 + beta^2 + 2alphabeta 8^2 = 34 + 2alphabeta implies 64 - 34 = 2alphabeta implies 2alphabeta = 30 implies alphabeta = 15 Since we know both the \sum (8) and product (15), alpha and \beta are the roots of the quadratic equation: x^2 - 8x + 15 = 0 (x - 3)(x - 5) = 0 implies x = 3, \, 5 ### Step 2: Assign Variables and Compute the Target Value Using the given constraint condition alpha leq beta, we assign the values as: alpha = 3, quad beta = 5 Now substitute these values into the evaluation expression: alpha^2 + beta = 3^2 + 5 = 9 + 5 = 14 ### Pattern Recognition Recognizing standard algebraic forms for sums and products like alpha+beta and alphabeta helps identify the system's values without needing to use full square root quadratic formulas. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Inverse Trigonometric Functions

Reference Study Guides

More Inverse Trigonometric Functions Previous-Year Questions — Page 3

Q15 jee_main_2024_31_jan_morning Properties of Inverse Trigonometric Functions
For alpha, beta, gamma neq 0. If sin^-1alpha + sin^-1beta + sin^-1gamma = pi and (alpha + beta + gamma)(alpha - gamma + beta) = 3 alphabeta then gamma equal to
  • A. fracsqrt32
  • B. frac1sqrt2
  • C. fracsqrt3 - 12sqrt2
  • D. sqrt3

Solution

### Core Logic Let sin^-1alpha = A, sin^-1beta = B, sin^-1gamma = C. Given A + B + C = pi. Since sin A = alpha, sin B = beta, sin C = gamma, alpha, beta, gamma act like the side lengths of a triangle divided by 2R by Sine rule. However, directly dealing with the relation: (alpha + beta + gamma)(alpha + beta - gamma) = 3alphabeta ### Step 1: Simplify Algebraic Relation (alpha + beta)^2 - gamma^2 = 3alphabeta alpha^2 + beta^2 + 2alphabeta - gamma^2 = 3alphabeta alpha^2 + beta^2 - gamma^2 = alphabeta ### Step 2: Triangle Identification Divide by 2alphabeta: fracalpha^2 + beta^2 - gamma^22alphabeta = frac12 By Cosine Rule, cos C = frac12. Since C = sin^-1gamma, we know sin C = gamma. cos C = sqrt1 - gamma^2 = frac12. ### Step 3: Final Solution 1 - gamma^2 = frac14 implies gamma^2 = frac34 Since C is an angle of a triangle (or sum equals pi and elements are positive limits), gamma = sin C > 0. gamma = fracsqrt32 ### Pattern Recognition The expression (alpha + beta + gamma)(alpha + beta - gamma) = 3alphabeta perfectly mirrors the Cosine Rule standard form giving cos C = 1/2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Trigonometric Functions

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