JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
Given the inverse trigonometric function assumes principal values only. Let \(\mathrm{x}, \mathrm{y}\) be any two real numbers in \([-1,1]\) such that \(\cos ^{-1} \mathrm{x}-\sin ^{-1} \mathrm{y}=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi \text {. }\) Then, the minimum value of \(x^2+y^2+2 x y \sin \alpha\) is
- A \(-1\)
- B \(0\)
- C \(\frac{-1}{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(0\)
Step-by-step Solution
Detailed explanation
\( \cos ^{-1} x-\left(\frac{\pi}{2}-\cos ^{-1} y\right)=\alpha \) \( \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \) \( \alpha \in\left[-\frac{\pi}{2}, \pi\right], \frac{\pi}{2}+\alpha \in\left[0, \frac{3 \pi}{2}\right] \)…
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