Considering the principal values of the inverse trigonometric functions, sin^-1left(fracsqrt32 x + frac12sqrt1 - x^2right)$\sin^{-1}\left(\frac{\sqrt{3}}{2} x + \frac{1}{2}\sqrt{1 - x^2}\right)$, where -frac12 < x < frac1sqrt2$-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$, is equal to

### Core Logic Let sin^-1alpha = A, sin^-1beta = B, sin^-1gamma = C$\sin^{-1}\alpha = A, \sin^{-1}\beta = B, \sin^{-1}\gamma = C$. Given A + B + C = pi$A + B + C = \pi$. Since sin A = alpha, sin B = beta, sin C = gamma$\sin A = \alpha, \sin B = \beta, \sin C = \gamma$, alpha, beta, gamma$\alpha, \beta, \gamma$ act like the side lengths of a triangle divided by 2R$2R$ by Sine rule. However, directly dealing with the relation: (alpha + beta + gamma)(alpha + beta - gamma) = 3alphabeta$$(\alpha + \beta + \gamma)(\alpha + \beta - \gamma) = 3\alpha\beta$$ ### Step 1: Simplify Algebraic Relation (alpha + beta)^2 - gamma^2 = 3alphabeta$$(\alpha + \beta)^2 - \gamma^2 = 3\alpha\beta$$ alpha^2 + beta^2 + 2alphabeta - gamma^2 = 3alphabeta$$\alpha^2 + \beta^2 + 2\alpha\beta - \gamma^2 = 3\alpha\beta$$ alpha^2 + beta^2 - gamma^2 = alphabeta$$\alpha^2 + \beta^2 - \gamma^2 = \alpha\beta$$ ### Step 2: Triangle Identification Divide by 2alphabeta$2\alpha\beta$: fracalpha^2 + beta^2 - gamma^22alphabeta = frac12$$\frac{\alpha^2 + \beta^2 - \gamma^2}{2\alpha\beta} = \frac{1}{2}$$ By Cosine Rule, cos C = frac12$\cos C = \frac{1}{2}$. Since C = sin^-1gamma$C = \sin^{-1}\gamma$, we know sin C = gamma$\sin C = \gamma$. cos C = sqrt1 - gamma^2 = frac12$\cos C = \sqrt{1 - \gamma^2} = \frac{1}{2}$. ### Step 3: Final Solution 1 - gamma^2 = frac14 implies gamma^2 = frac34$$1 - \gamma^2 = \frac{1}{4} \implies \gamma^2 = \frac{3}{4}$$ Since C$C$ is an angle of a triangle (or sum equals pi$\pi$ and elements are positive limits), gamma = sin C > 0$\gamma = \sin C > 0$. gamma = fracsqrt32$$\gamma = \frac{\sqrt{3}}{2}$$ ### Pattern Recognition The expression (alpha + beta + gamma)(alpha + beta - gamma) = 3alphabeta$(\alpha + \beta + \gamma)(\alpha + \beta - \gamma) = 3\alpha\beta$ perfectly mirrors the Cosine Rule standard form giving cos C = 1/2$\cos C = 1/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Trigonometric Functions