Given that the inverse trigonometric function assumes principal values only. Let be any two real numbers in such that Then, the minimum value of is

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Solution

Q.
Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[- 1, 1]$ such that $\cos^{- 1} x - \sin^{- 1} y = α,$ $\frac{- π}{2} \leq α \leq π .$ Then, the minimum value of $x^{2} + y^{2} + 2 x y$ $\sin α$ is [2024]

1 $- 1$

2 $0$

3 $\frac{1}{2}$

4 $\frac{- 1}{2}$

Ans.
(2)

$\cos^{- 1} x - \frac{π}{2} + \cos^{- 1} y = α$

$\Rightarrow \cos^{- 1} x + \cos^{- 1} y = \frac{π}{2} + α$

$∴$ $\frac{π}{2} + α \in (0, \frac{3 π}{2})$ $[∵ α \in [\frac{- π}{2}, π]]$

$\cos^{- 1} (x y - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}) = \frac{π}{2} + α$

$\Rightarrow x y - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} = - \sin α$

$\Rightarrow x y + \sin α = \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}$

$\Rightarrow x^{2} y^{2} + \sin^{2} α + 2 x y \sin α = 1 - x^{2} - y^{2} + x^{2} y^{2}$

$\Rightarrow x^{2} + y^{2} + 2 x y \sin α = \cos^{2} α$

Thus, required minimum value is 0.

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