If y = cos left( fracpi3 + cos^-1 fracx2 right)$y = \cos \left( \frac{\pi}{3} + \cos^{-1} \frac{x}{2} \right)$, then (x - y)^2 + 3y^2$(x - y)^2 + 3y^2$ is equal to ____________.

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

### Related Formula Apply the fundamental trigonometric identity properties directly linking reciprocal functions: sec^2(theta) = 1 + tan^2(theta)$$\sec^2(\theta) = 1 + \tan^2(\theta)$$ csc^2(phi) = 1 + cot^2(phi)$$\csc^2(\phi) = 1 + \cot^2(\phi)$$ ### Core Logic Simplify the given trigonometric equation using the identity formulas: sec^2(tan^-1alpha) = 1 + tan^2(tan^-1alpha) = 1 + alpha^2$$\sec^2(\tan^{-1}\alpha) = 1 + \tan^2(\tan^{-1}\alpha) = 1 + \alpha^2$$ csc^2(cot^-1beta) = 1 + cot^2(cot^-1beta) = 1 + beta^2$$\csc^2(\cot^{-1}\beta) = 1 + \cot^2(\cot^{-1}\beta) = 1 + \beta^2$$ Substitute these simplified expressions back into the target relation equation: (1 + alpha^2) + (1 + beta^2) = 36 implies alpha^2 + beta^2 = 34$$(1 + \alpha^2) + (1 + \beta^2) = 36 \implies \alpha^2 + \beta^2 = 34$$ ### Step 1: Set up a Quadratic Equation for the roots We are given the linear \sum alpha + beta = 8$\alpha + \beta = 8$. Use the algebraic identity for squares to find the product: (alpha + beta)^2 = alpha^2 + beta^2 + 2alphabeta$$(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$$ 8^2 = 34 + 2alphabeta implies 64 - 34 = 2alphabeta implies 2alphabeta = 30 implies alphabeta = 15$$8^2 = 34 + 2\alpha\beta \implies 64 - 34 = 2\alpha\beta \implies 2\alpha\beta = 30 \implies \alpha\beta = 15$$ Since we know both the \sum (8) and product (15), alpha$\alpha$ and \beta$\beta$ are the roots of the quadratic equation: x^2 - 8x + 15 = 0$$x^2 - 8x + 15 = 0$$ (x - 3)(x - 5) = 0 implies x = 3, \, 5$$(x - 3)(x - 5) = 0 \implies x = 3, \, 5$$ ### Step 2: Assign Variables and Compute the Target Value Using the given constraint condition alpha leq beta$\alpha \leq \beta$, we assign the values as: alpha = 3, quad beta = 5$$\alpha = 3, \quad \beta = 5$$ Now substitute these values into the evaluation expression: alpha^2 + beta = 3^2 + 5 = 9 + 5 = 14$$\alpha^2 + \beta = 3^2 + 5 = 9 + 5 = 14$$ ### Pattern Recognition Recognizing standard algebraic forms for sums and products like alpha+beta$\alpha+\beta$ and alphabeta$\alpha\beta$ 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

### Related Formula sin^-1 x + cos^-1 x = fracpi2$$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$$ cos 2alpha = 2cos^2 alpha - 1$$\cos 2\alpha = 2\cos^2 \alpha - 1$$ ### Core Logic Rearrange using the principal trigonometric identity property sin^-1 x = fracpi2 - cos^-1 x$\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$: cos^-1 x = pi + left(fracpi2 - cos^-1 xright) + sin^-1(2x + 1)$$\cos^{-1} x = \pi + \left(\frac{\pi}{2} - \cos^{-1} x\right) + \sin^{-1}(2x + 1)$$ 2cos^-1 x - sin^-1(2x + 1) = frac3pi2$$2\cos^{-1} x - \sin^{-1}(2x + 1) = \frac{3\pi}{2}$$ ### Step 1: Isolate angles using variable substitution Let cos^-1 x = alpha$\cos^{-1} x = \alpha$ and sin^-1(2x + 1) = beta$\sin^{-1}(2x + 1) = \beta$. 2alpha - beta = frac3pi2 implies 2alpha = frac3pi2 + beta$$2\alpha - \beta = \frac{3\pi}{2} \implies 2\alpha = \frac{3\pi}{2} + \beta$$ Take cosine on both sides: cos(2alpha) = cosleft(frac3pi2 + betaright) = sinbeta$$\cos(2\alpha) = \cos\left(\frac{3\pi}{2} + \beta\right) = \sin\beta$$ ### Step 2: Convert to algebraic identity form Using double \angle formulas: 2cos^2 alpha - 1 = sinbeta$$2\cos^2 \alpha - 1 = \sin\beta$$ Since cosalpha = x$\cos\alpha = x$ and sinbeta = 2x + 1$\sin\beta = 2x + 1$, substitute directly: 2x^2 - 1 = 2x + 1 implies 2x^2 - 2x - 2 = 0 implies x^2 - x - 1 = 0$$2x^2 - 1 = 2x + 1 \implies 2x^2 - 2x - 2 = 0 \implies x^2 - x - 1 = 0$$ Solving the quadratic equation gives: x = frac1 pm sqrt52$$x = \frac{1 \pm \sqrt{5}}{2}$$ Checking domain validation constraints for inverse functions dictates that only x = frac1 - sqrt52$x = \frac{1 - \sqrt{5}}{2}$ is valid (the positive root exceeds principal bounds). ### Step 3: Evaluate target question calculation From the valid root, we track: 2x - 1 = -sqrt5$$2x - 1 = -\sqrt{5}$$ Squaring both sides yields: (2x - 1)^2 = (-sqrt5)^2 = 5$$(2x - 1)^2 = (-\sqrt{5})^2 = 5$$ ### Pattern Recognition When inverse sums equate to values like frac3pi2$\frac{3\pi}{2}$, 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

### Related Formula cos(2theta) = 1 - 2sin^2theta$\cos(2\theta) = 1 - 2\sin^2\theta$ ### Core Logic Let sin^-1x = theta implies sintheta = x$\sin^{-1}x = \theta \implies \sin\theta = x$. The equation matches cos(2theta) = frac19$\cos(2\theta) = \frac{1}{9}$: 1 - 2sin^2theta = frac19 implies 1 - 2x^2 = frac19$$1 - 2\sin^2\theta = \frac{1}{9} \implies 1 - 2x^2 = \frac{1}{9}$$ 2x^2 = 1 - frac19 = frac89 implies x^2 = frac49 implies x = pm frac23$$2x^2 = 1 - \frac{1}{9} = \frac{8}{9} \implies x^2 = \frac{4}{9} \implies x = \pm \frac{2}{3}$$ Since m$m$ and n$n$ are natural numbers, we pick x = frac23 = fracmn$x = \frac{2}{3} = \frac{m}{n}$. Because 2$2$ and 3$3$ are co-prime, we choose m = 2$m = 2$ and n = 3$n = 3$. ### Step 1: Formulating Quadratic Equations Substituting values into mx^2 - nx - m + n = 0$mx^2 - nx - m + n = 0$: 2x^2 - 3x - 2 + 3 = 0 implies 2x^2 - 3x + 1 = 0$$2x^2 - 3x - 2 + 3 = 0 \implies 2x^2 - 3x + 1 = 0$$ Factoring the equations: (2x - 1)(x - 1) = 0 implies x = 1 text or x = frac12$$(2x - 1)(x - 1) = 0 \implies x = 1 \text{ or } x = \frac{1}{2}$$ Given alpha > beta$\alpha > \beta$, we have alpha = 1$\alpha = 1$ and beta = frac12$\beta = \frac{1}{2}$. ### Step 2: Checking Options Let us check the coordinates left(1, frac12right)$\left(1, \frac{1}{2}\right)$ against option line configurations: 5(1) + 8left(frac12right) = 5 + 4 = 9$$5(1) + 8\left(\frac{1}{2}\right) = 5 + 4 = 9$$ This exactly matches option (4). ### Pattern Recognition Co-prime conditions uniquely lock fractional values down to absolute integers. This bridges variables directly into standard algebraic calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Inverse Trigonometric Functions Class 10 Mathematics: Quadratic Equations

### Related Formula sin^-1(sin x) = x - 2pi text for x in [3pi/2, 5pi/2]$$\sin^{-1}(\sin x) = x - 2\pi \text{ for } x \in [3\pi/2, 5\pi/2]$$ cos^-1(cos x) = 2pi - x text for x in [pi, 2pi]$$\cos^{-1}(\cos x) = 2\pi - x \text{ for } x \in [\pi, 2\pi]$$ ### Core Logic Evaluate a = sin^-1(sin 5)$a = \sin^{-1}(\sin 5)$: The principal branch of sin^-1 x$\sin^{-1} x$ is [-pi/2, pi/2]$[-\pi/2, \pi/2]$. 5$5$ radians is approximately 5 times 57.3^circ approx 286.5^circ$5 \times 57.3^\circ \approx 286.5^\circ$ (in 4th quadrant). The equivalent angle in the principal domain is 5 - 2pi$5 - 2\pi$. Thus, a = 5 - 2pi$a = 5 - 2\pi$. Evaluate b = cos^-1(cos 5)$b = \cos^{-1}(\cos 5)$: The principal branch of cos^-1 x$\cos^{-1} x$ is [0, pi]$[0, \pi]$. 5$5$ radians is in [pi, 2pi]$[\pi, 2\pi]$. The equivalent angle is 2pi - 5$2\pi - 5$. Thus, b = 2pi - 5$b = 2\pi - 5$. Calculate a^2 + b^2$a^2 + b^2$: a^2 + b^2 = (5 - 2pi)^2 + (2pi - 5)^2$$a^2 + b^2 = (5 - 2\pi)^2 + (2\pi - 5)^2$$ = 2(5 - 2pi)^2$= 2(5 - 2\pi)^2$ = 2(25 + 4pi^2 - 20pi)$$= 2(25 + 4\pi^2 - 20\pi)$$ = 8pi^2 - 40pi + 50$$= 8\pi^2 - 40\pi + 50$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Inverse Trigonometric Functions