If alpha>beta>gamma>0$\alpha>\beta>\gamma>0$ 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\$\cot^{-1}\left\{\beta+\frac{(1+\beta^{2})}{(\alpha-\beta)}\right\}+\cot^{-1}\left\{\gamma+\frac{(1+\gamma^{2})}{(\beta-\gamma)}\right\}+\cot^{-1}\left\{\alpha+\frac{(1+\alpha^{2})}{(\gamma-\alpha)}\right\}$ is equal to: [cite: 3257, 3258]

### Related Formula cos(A + B) = cos A cos B - sin A sin B$$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ sinleft(cos^-1 uright) = sqrt1 - u^2$$\sin\left(\cos^{-1} u\right) = \sqrt{1 - u^2}$$ ### Core Logic We expand the trigonometric compound angle expression to obtain a coupled algebraic equation relating variables x$x$ and y$y$. ### Step 1: Expand the equation using cosine addition formula Let theta = cos^-1left(fracx2right) implies cos theta = fracx2$\theta = \cos^{-1}\left(\frac{x}{2}\right) \implies \cos \theta = \frac{x}{2}$ and sin theta = sqrt1 - fracx^24$\sin \theta = \sqrt{1 - \frac{x^2}{4}}$: y = cosleft(fracpi3 + thetaright) = cosfracpi3 costheta - sinfracpi3 sintheta$$y = \cos\left(\frac{\pi}{3} + \theta\right) = \cos\frac{\pi}{3} \cos\theta - \sin\frac{\pi}{3} \sin\theta$$ y = frac12 left( fracx2 right) - fracsqrt32 sqrt1 - fracx^24$$y = \frac{1}{2} \left( \frac{x}{2} \right) - \frac{\sqrt{3}}{2} \sqrt{1 - \frac{x^2}{4}}$$ y = fracx4 - fracsqrt34 sqrt4 - x^2$$y = \frac{x}{4} - \frac{\sqrt{3}}{4} \sqrt{4 - x^2}$$ 4y = x - sqrt3sqrt4 - x^2$$4y = x - \sqrt{3}\sqrt{4 - x^2}$$ ### Step 2: Isolate the root and square Rearrange terms to isolate the radical and square both sides: x - 4y = sqrt3sqrt4 - x^2$$x - 4y = \sqrt{3}\sqrt{4 - x^2}$$ (x - 4y)^2 = 3(4 - x^2)$$(x - 4y)^2 = 3(4 - x^2)$$ x^2 - 8xy + 16y^2 = 12 - 3x^2$$x^2 - 8xy + 16y^2 = 12 - 3x^2$$ 4x^2 - 8xy + 16y^2 = 12$$4x^2 - 8xy + 16y^2 = 12$$ Divide the entire equation by 4: x^2 - 2xy + 4y^2 = 3$$x^2 - 2xy + 4y^2 = 3$$ ### Step 3: Evaluate the target expression We want to find the value of (x-y)^2 + 3y^2$(x-y)^2 + 3y^2$: (x-y)^2 + 3y^2 = x^2 - 2xy + y^2 + 3y^2 = x^2 - 2xy + 4y^2$$(x-y)^2 + 3y^2 = x^2 - 2xy + y^2 + 3y^2 = x^2 - 2xy + 4y^2$$ Notice that this matches the left side of our simplified equation from Step 2 exactly: (x - y)^2 + 3y^2 = 3$$(x - y)^2 + 3y^2 = 3$$ ### Pattern Recognition Coefficient symmetry: The expression (x-y)^2 + 3y^2 = x^2 - 2xy + 4y^2$(x-y)^2 + 3y^2 = x^2 - 2xy + 4y^2$ is a standard algebraic representation designed to match the quadratic expansion of scaled trigonometric sum equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Inverse Trigonometric Functions

### Related Formula sqrt1+tan^2theta = |sectheta|$$\sqrt{1+\tan^2\theta} = |\sec\theta|$$ ### Core Logic Track angular positions across quadrants accurately. Evaluate positive/negative absolute value signs based on component radian locations before reducing formulas. ### Step 1: Simplify First Exponent Operand For tracking segment angle height theta = 2$\theta = 2$ radians (Quadrant II), cosine terms switch below zero: |sec 2| = -sec 2$$|\sec 2| = -\sec 2$$ frac-sec 2 - 1tan 2 = frac-1 - cos 2sin 2 = -cot 1$$\frac{-\sec 2 - 1}{\tan 2} = \frac{-1 - \cos 2}{\sin 2} = -\cot 1$$ ### Step 2: Simplify Second Exponent Operand For tracking segment angle height theta = 1/2$\theta = 1/2$ radian (Quadrant I), expressions remain positive: |sec(1/2)| = sec(1/2)$$|\sec(1/2)| = \sec(1/2)$$ fracsec(1/2) + 1tan(1/2) = frac1 + cos(1/2)sin(1/2) = cot(1/4)$$\frac{\sec(1/2) + 1}{\tan(1/2)} = \frac{1 + \cos(1/2)}{\sin(1/2)} = \cot(1/4)$$ ### Step 3: Combine Structural Terms Apply inverse mapping functions carefully: cot^-1(-cot 1) - cot^-1left(cot frac14right) = (pi - 1) - frac14 = pi - frac54$$\cot^{-1}(-\cot 1) - \cot^{-1}\left(\cot \frac{1}{4}\right) = (\pi - 1) - \frac{1}{4} = \pi - \frac{5}{4}$$ ### Pattern Recognition Radian values like 2 sit over 90^circ$90^\circ$ but beneath 180^circ$180^\circ$. Missing quadrant validation tags is a common trap in inverse identity tracking. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Inverse Trigonometric Functions

### Related Formula Standard tangent identity sum format: tan^-1 x + tan^-1 y = tan^-1 left(fracx+y1-xyright)$$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\frac{x+y}{1-xy}\right)$$ ### Core Logic Convert all components into tangent mappings: sin^-1frac35 = tan^-1frac34$\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{3}{4}$ sin^-1frac513 = tan^-1frac512$\sin^{-1}\frac{5}{13} = \tan^{-1}\frac{5}{12}$ sin^-1frac3365 = tan^-1frac3356$\sin^{-1}\frac{33}{65} = \tan^{-1}\frac{33}{56}$ ### Step 1: Evaluating the Mapped Component Sum Summing the first two components: tan^-1frac34 + tan^-1frac512 = tan^-1left(fracfrac34 + frac5121 - frac1548right) = tan^-1frac5633$$\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12} = \tan^{-1}\left(\frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{15}{48}}\right) = \tan^{-1}\frac{56}{33}$$ ### Step 2: Applying Cofunction Complements Notice that tan^-1frac3356 = cot^-1frac5633$\tan^{-1}\frac{33}{56} = \cot^{-1}\frac{56}{33}$. Combining everything inside the function: cos left(tan^-1frac5633 + cot^-1frac5633right) = cos left(fracpi2 ight) = 0$$\cos \left(\tan^{-1}\frac{56}{33} + \cot^{-1}\frac{56}{33}\right) = \cos \left(\frac{\pi}{2} ight) = 0$$ ### Pattern Recognition Look for reciprocal fractional identities across matching inverse blocks—they easily merge using the tan^-1 x + cot^-1 x = fracpi2$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ identity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Inverse Trigonometric Functions

### Related Formula The general transformation formula for a difference of tangents is: tan^-1x - tan^-1y = tan^-1left(fracx-y1+xyright)$$\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$$ ### Core Logic Let the general term of the series be T_n$T_n$. Examining the numerators (7, 19, 39, 67, dots$7, 19, 39, 67, \dots$): The differences between consecutive terms are 12, 20, 28, dots$12, 20, 28, \dots$, which forms an arithmetic progression with a common difference of 8$8$. Thus, the general term for the sequence in the numerator can be found using difference methods: textNumerator = 4n^2 + 3$$\text{Numerator } = 4n^2 + 3$$ Therefore, the n$n$-th term T_n$T_n$ is: T_n = cot^-1left(frac4n^2 + 34right) = tan^-1left(frac44n^2 + 3right)$$T_n = \cot^{-1}\left(\frac{4n^2 + 3}{4}\right) = \tan^{-1}\left(\frac{4}{4n^2 + 3}\right)$$ ### Step 1: Rewriting the general term for telescoping sum Divide the numerator and denominator inside the argument by 4$4$: T_n = tan^-1left(frac1n^2 + frac34right) = tan^-1left(frac11 + left(n^2 - frac14right)right)$$T_n = \tan^{-1}\left(\frac{1}{n^2 + \frac{3}{4}}\right) = \tan^{-1}\left(\frac{1}{1 + \left(n^2 - \frac{1}{4}\right)}\right)$$ Factorize n^2 - frac14$n^2 - \frac{1}{4}$ as a difference of squares: T_n = tan^-1left(fracleft(n + frac12right) - left(n - frac12right)1 + left(n + frac12right)left(n - frac12right)right)$$T_n = \tan^{-1}\left(\frac{\left(n + \frac{1}{2}\right) - \left(n - \frac{1}{2}\right)}{1 + \left(n + \frac{1}{2}\right)\left(n - \frac{1}{2}\right)}\right)$$ Using the difference formula for tan^-1$\tan^{-1}$: T_n = tan^-1left(n + frac12right) - tan^-1left(n - frac12right)$$T_n = \tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(n - \frac{1}{2}\right)$$ ### Step 2: Telescoping summation Expanding the sum up to n$n$ terms: S_n = sum_k=1^n T_k = left[tan^-1left(frac32right) - tan^-1left(frac12right)right] + left[tan^-1left(frac52right) - tan^-1left(frac32right)right] + dots + left[tan^-1left(n + frac12right) - tan^-1left(n - frac12right)right]$$S_n = \sum_{k=1}^n T_k = \left[\tan^{-1}\left(\frac{3}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right)\right] + \left[\tan^{-1}\left(\frac{5}{2}\right) - \tan^{-1}\left(\frac{3}{2}\right)\right] + \dots + \left[\tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(n - \frac{1}{2}\right)\right]$$ All intermediate terms cancel out, leaving: S_n = tan^-1left(n + frac12right) - tan^-1left(frac12right)$$S_n = \tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right)$$ ### Step 3: Infinite limit evaluation Taking the limit as n to infty$n \to \infty$: S_infty = lim_n to infty left[tan^-1left(n + frac12right) - tan^-1left(frac12right)right] = fracpi2 - tan^-1left(frac12right)$$S_\infty = \lim_{n \to \infty} \left[\tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right)\right] = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{2}\right)$$ ### Pattern Recognition Whenever you see an infinite series involving cot^-1$\cot^{-1}$ or tan^-1$\tan^{-1}$, try to rearrange the denominator into the form 1 + xy$1 + xy$ and check if the numerator matches x - y$x - y$ to set up a standard telescoping structure. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Inverse Trigonometric Functions Class 11 Mathematics: Sequences and Series

### Related Formula Trigonometric Sine Identity: sin(A + B) = sin A cos B + cos A sin B$$\sin(A + B) = \sin A \cos B + \cos A \sin B$$ ### Core Logic Let sin^-1x = theta implies x = sintheta$\sin^{-1}x = \theta \implies x = \sin\theta$ and sqrt1-x^2 = costheta$\sqrt{1-x^2} = \cos\theta$. Given constraint -frac12 < x < frac1sqrt2 implies -fracpi6 < theta < fracpi4$-\frac{1}{2} < x < \frac{1}{\sqrt{2}} \implies -\frac{\pi}{6} < \theta < \frac{\pi}{4}$. Substitute parameter representations into expression: sin^-1left(fracsqrt32sintheta + frac12costhetaright) = sin^-1left(sinthetacosfracpi6 + costhetasinfracpi6right)$$\sin^{-1}\left(\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta\right) = \sin^{-1}\left(\sin\theta\cos\frac{\pi}{6} + \cos\theta\sin\frac{\pi}{6}\right)$$ sin^-1left(sinleft(theta + fracpi6right)right)$$\sin^{-1}\left(\sin\left(\theta + \frac{\pi}{6}\right)\right)$$ ### Step 1: Check Principal Bounds Evaluate bounds for arguments: since -fracpi6 < theta < fracpi4$-\frac{\pi}{6} < \theta < \frac{\pi}{4}$: -fracpi6 + fracpi6 < theta + fracpi6 < fracpi4 + fracpi6 implies 0 < theta + fracpi6 < frac5pi12$$-\frac{\pi}{6} + \frac{\pi}{6} < \theta + \frac{\pi}{6} < \frac{\pi}{4} + \frac{\pi}{6} \implies 0 < \theta + \frac{\pi}{6} < \frac{5\pi}{12}$$ This lies completely within the principal value branch of sin^-1x$\sin^{-1}x$, which is left[-fracpi2, fracpi2right]$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, sin^-1left(sinleft(theta + fracpi6right)right) = theta + fracpi6$\sin^{-1}\left(\sin\left(\theta + \frac{\pi}{6}\right)\right) = \theta + \frac{\pi}{6}$. ### Step 2: Final Form Substituting back theta = sin^-1x$\theta = \sin^{-1}x$: fracpi6 + sin^-1x$$\frac{\pi}{6} + \sin^{-1}x$$ ### Pattern Recognition Always check primary interval bounds when stripping inverse operators. If the arguments exceed bounds, quadrant mapping transformations must be performed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Inverse Trigonometric Functions