sin ^{-1} frac{sqrt{3}}{2} + sin ^{-1} sqrt{f | vedclass

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(B) माना कि $x = \frac{\sqrt{3}}{2}$ और $y = \sqrt{\frac{2}{3}}$.ध्यान दें कि $x^2 + y^2 = \frac{3}{4} + \frac{2}{3} = \frac{9+8}{12} = \frac{17}{12}

Vedclass · Accepted Answer

$\pi - \sin ^{-1} \left( \frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}} \right)$

Vedclass · Answer

(B) माना कि $x = \frac{\sqrt{3}}{2}$ और $y = \sqrt{\frac{2}{3}}$.ध्यान दें कि $x^2 + y^2 = \frac{3}{4} + \frac{2}{3} = \frac{9+8}{12} = \frac{17}{12} > 1$.चूंकि $x, y > 0$ और $x^2 + y^2 > 1$, हम सर्वसमिका $\sin^{-1} x + \sin^{-1} y = \pi - \sin^{-1} (x \sqrt{1-y^2} + y \sqrt{1-x^2})$ का उपयोग करते हैं।मान रखने पर:$\sin^{-1} \frac{\sqrt{3}}{2} + \sin^{-1} \sqrt{\frac{2}{3}} = \pi - \sin^{-1} \left( \frac{\sqrt{3}}{2} \sqrt{1 - \frac{2}{3}} + \sqrt{\frac{2}{3}} \sqrt{1 - \frac{3}{4}} \right)$$= \pi - \sin^{-1} \left( \frac{\sqrt{3}}{2} \sqrt{\frac{1}{3}} + \sqrt{\frac{2}{3}} \sqrt{\frac{1}{4}} \right)$$= \pi - \sin^{-1} \left( \frac{\sqrt{3}}{2 \sqrt{3}} + \frac{\sqrt{2}}{\sqrt{3} \cdot 2} \right)$$= \pi - \sin^{-1} \left( \frac{1}{2} + \frac{\sqrt{2}}{2 \sqrt{3}} \right)$$= \pi - \sin^{-1} \left( \frac{\sqrt{3} + \sqrt{2}}{2 \sqrt{3}} \right)$.

$\sin ^{-1} \frac{\sqrt{3}}{2} + \sin ^{-1} \sqrt{\frac{2}{3}} = $

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