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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

Inverse Trigonometric Functions

More Inverse Trigonometric Functions Previous-Year Questions — Page 2

Q53 2025 Properties of Inverse Trigonometric Functions
If alpha>beta>gamma>0 then the expression cot^-1left\beta+frac(1+beta^2)(alpha-beta)right\+cot^-1left\gamma+frac(1+gamma^2)(beta-gamma)right\+cot^-1left\alpha+frac(1+alpha^2)(gamma-alpha)right\ is equal to: [cite: 3257, 3258]
  • A. fracpi2-(alpha+beta+gamma)
  • B. 3pi
  • C. 0
  • D. pi

Solution

### Related Formula The standard conversion between cot^-1(x) and tan^-1(x) depends on the sign of x: cot^-1(x) = tan^-1left(frac1xright) quad textif x > 0 cot^-1(x) = pi + tan^-1left(frac1xright) quad textif x < 0 ### Core Logic Simplify the interior terms algebraic representations: beta + frac1+beta^2alpha-beta = fracalphabeta - beta^2 + 1 + beta^2alpha-beta = frac1+alphabetaalpha-beta gamma + frac1+gamma^2beta-gamma = fracbetagamma - gamma^2 + 1 + gamma^2beta-gamma = frac1+betagammabeta-gamma alpha + frac1+alpha^2gamma-alpha = fracalphagamma - alpha^2 + 1 + alpha^2gamma-alpha = frac1+alphagammagamma-alpha ### Step 1: Convert to Inverse Tangent terms Since alpha > beta > gamma > 0: 1. frac1+alphabetaalpha-beta > 0 Rightarrow cot^-1left(frac1+alphabetaalpha-betaright) = tan^-1left(fracalpha-beta1+alphabetaright) 2. frac1+betagammabeta-gamma > 0 Rightarrow cot^-1left(frac1+betagammabeta-gammaright) = tan^-1left(fracbeta-gamma1+betagammaright) 3. frac1+alphagammagamma-alpha < 0 (since gamma - alpha < 0) Rightarrow cot^-1left(frac1+alphagammagamma-alpharight) = pi + tan^-1left(fracgamma-alpha1+alphagammaright) ### Step 2: Telescopic Sum Evaluation Apply the difference identity for arctan, tan^-1left(fracx-y1+xyright) = tan^-1x - tan^-1y : = (tan^-1alpha - tan^-1beta) + (tan^-1beta - tan^-1gamma) + pi + (tan^-1gamma - tan^-1alpha) All variables cancel symmetrically leaving[cite: 3886, 3887]: = pi ### Pattern Recognition The sign trap is the most vital component of this question. The ordering alpha > beta > gamma > 0 means the last term contains a denominator with a negative difference (gamma - alpha), introducing the +pi offset according to the principal range of cot^-1(x). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Inverse Trigonometric Functions
Q75 2025 Inverse Trigonometric Equations
Let S = left\x:cos^-1x = pi +sin^-1x + sin^-1left(2x + 1right)right\. Then sum_xin Sleft(2x - 1right)^2 is equal to
Numerical Answer. Answer: 5

Solution

### Related Formula sin^-1 x + cos^-1 x = fracpi2 cos 2alpha = 2cos^2 alpha - 1 ### Core Logic Rearrange using the principal trigonometric identity property sin^-1 x = fracpi2 - cos^-1 x: cos^-1 x = pi + left(fracpi2 - cos^-1 xright) + sin^-1(2x + 1) 2cos^-1 x - sin^-1(2x + 1) = frac3pi2 ### Step 1: Isolate angles using variable substitution Let cos^-1 x = alpha and sin^-1(2x + 1) = beta. 2alpha - beta = frac3pi2 implies 2alpha = frac3pi2 + beta Take cosine on both sides: cos(2alpha) = cosleft(frac3pi2 + betaright) = sinbeta ### Step 2: Convert to algebraic identity form Using double \angle formulas: 2cos^2 alpha - 1 = sinbeta Since cosalpha = x and sinbeta = 2x + 1, substitute directly: 2x^2 - 1 = 2x + 1 implies 2x^2 - 2x - 2 = 0 implies x^2 - x - 1 = 0 Solving the quadratic equation gives: x = frac1 pm sqrt52 Checking domain validation constraints for inverse functions dictates that only x = frac1 - sqrt52 is valid (the positive root exceeds principal bounds). ### Step 3: Evaluate target question calculation From the valid root, we track: 2x - 1 = -sqrt5 Squaring both sides yields: (2x - 1)^2 = (-sqrt5)^2 = 5 ### Pattern Recognition When inverse sums equate to values like frac3pi2, look for extreme boundary values or domain constraints to eliminate invalid algebraic roots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Inverse Trigonometric Functions Class 11 Mathematics: Trigonometric Equations

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