For $\alpha, \beta, \gamma \neq 0$, if $\sin^{-1}\alpha+\sin^{-1}\beta+\sin^{-1}\gamma=\pi$ and $(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3\alpha\beta$, then $\gamma$ equals

(C) $\frac{\sqrt{3}-1}{2\sqrt{2}}$

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Inverse Trigonometric Functions

For

α, β, γ \neq = 0

, if

sin^{- 1} α + sin^{- 1} β + sin^{- 1} γ = π

and

(α + β + γ) (α - γ + β) = 3 α β

, then

γ

equals

Answer: (A)

Let

A = sin^{- 1} α, B = sin^{- 1} β, C = sin^{- 1} γ

, so

A + B + C = π

and

α = sin A, β = sin B, γ = sin C

.

Expanding the given condition:

(α + β + γ) (α + β - γ) = 3 α β ⟹ (α + β)^{2} - γ^{2} = 3 α β ⟹ α^{2} + β^{2} - α β = γ^{2}

.

Since

A, B, C

are angles of a triangle (they sum to

π

) with

α, β, γ

playing the role of the sides opposite them (as

sin A : sin B : sin C = α : β : γ

by construction, matching the law of sines for a triangle with circumdiameter 1), the law of cosines gives

γ^{2} = α^{2} + β^{2} - 2 α β cos C

Comparing with

γ^{2} = α^{2} + β^{2} - α β

gives

2 cos C = 1 ⟹ cos C = \frac{1}{2} ⟹ C = \frac{π}{3}

.

So

γ = sin C = sin \frac{π}{3} = \frac{3}{2}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option