If α>β>γ>0$\alpha>\beta>\gamma>0$ then the expression ⁻¹β+ (1+β²)(α-β)+ ⁻¹γ+ (1+γ²)(β-γ)+ ⁻¹α+ (1+α²)(γ-α)$\cot^{-1}\left\{\beta+\frac{(1+\beta^{2})}{(\alpha-\beta)}\right\}+\cot^{-1}\left\{\gamma+\frac{(1+\gamma^{2})}{(\beta-\gamma)}\right\}+\cot^{-1}\left\{\alpha+\frac{(1+\alpha^{2})}{(\gamma-\alpha)}\right\}$ is equal to:

Core Logic

Let ex = t$e^{\sin x} = t$, where t > 0$t > 0$ because exponential functions are strictly positive. Substitute into the equation:

Solve for t$t$ using the quadratic formula:

Since t > 0$t > 0$, we discard 1 - √(3)$1 - \sqrt{3}$. Thus, t = 1 + √(3) ≈ 2.732$t = 1 + \sqrt{3} \approx 2.732$. Now, equate back:

We know e ≈ 2.718$e \approx 2.718$. Since 1 + √(3) > e$1 + \sqrt{3} > e$, it follows that ln(1 + √(3)) > 1$\ln(1 + \sqrt{3}) > 1$. But the range of x$\sin x$ is [-1, 1]$[-1, 1]$. Therefore, x$\sin x$ cannot equal a value strictly greater than 1. No real solution exists.

Chapter Mix

Class 11 Maths: Trigonometric Functions Class 12 Maths: Continuity and Differentiability

Related Formula

Core Logic

Evaluate a = ⁻¹( 5)$a = \sin^{-1}(\sin 5)$: The principal branch of ⁻¹ x$\sin^{-1} x$ is [-π/2, π/2]$[-\pi/2, \pi/2]$. 5$5$ radians is approximately 5 × 57.3^° ≈ 286.5^°$5 \times 57.3^\circ \approx 286.5^\circ$ (in 4th quadrant). The equivalent angle in the principal domain is 5 - 2π$5 - 2\pi$. Thus, a = 5 - 2π$a = 5 - 2\pi$.

Evaluate b = ⁻¹( 5)$b = \cos^{-1}(\cos 5)$: The principal branch of ⁻¹ x$\cos^{-1} x$ is [0, π]$[0, \pi]$. 5$5$ radians is in [π, 2π]$[\pi, 2\pi]$. The equivalent angle is 2π - 5$2\pi - 5$. Thus, b = 2π - 5$b = 2\pi - 5$.

Calculate a² + b²$a^2 + b^2$:

= 2(5 - 2π)²$= 2(5 - 2\pi)^2$

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions

Core Logic

Let ⁻¹α = A, ⁻¹β = B, ⁻¹γ = C$\sin^{-1}\alpha = A, \sin^{-1}\beta = B, \sin^{-1}\gamma = C$. Given A + B + C = π$A + B + C = \pi$. Since A = α, B = β, C = γ$\sin A = \alpha, \sin B = \beta, \sin C = \gamma$, α, β, γ$\alpha, \beta, \gamma$ act like the side lengths of a triangle divided by 2R$2R$ by Sine rule. However, directly dealing with the relation:

Step 1: Simplify Algebraic Relation

Step 2: Triangle Identification

Divide by 2αβ$2\alpha\beta$:

By Cosine Rule, C = (1)/(2)$\cos C = \frac{1}{2}$. Since C = ⁻¹γ$C = \sin^{-1}\gamma$, we know C = γ$\sin C = \gamma$. C = √(1 - γ²) = (1)/(2)$\cos C = \sqrt{1 - \gamma^2} = \frac{1}{2}$.

Step 3: Final Solution

Since C$C$ is an angle of a triangle (or sum equals π$\pi$ and elements are positive limits), γ = C > 0$\gamma = \sin C > 0$.

Pattern Recognition

The expression (α + β + γ)(α + β - γ) = 3αβ$(\alpha + \beta + \gamma)(\alpha + \beta - \gamma) = 3\alpha\beta$ perfectly mirrors the Cosine Rule standard form giving C = 1/2$\cos C = 1/2$.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Trigonometric Functions