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}} = $

Similar Questions

यदि $f(n) = \tan \left[\tan ^{-1} \frac{1}{1+2} + \tan ^{-1} \frac{1}{1+6} + \tan ^{-1} \frac{1}{1+12} + \ldots + \tan ^{-1} \frac{1}{1+n(n+1)}\right]$ है, तो $f(2021) =$

$\cos \left(\tan ^{-1}\left(\sin \left(\cot ^{-1} x\right)\right)\right)$ का मान है

$2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=$

सिद्ध कीजिए कि $\cos ^{-1} \frac{4}{5} + \cos ^{-1} \frac{12}{13} = \cos ^{-1} \frac{33}{65}$

यदि $A=2 \tan ^{-1}\left(\frac{1+x}{1-x}\right)$ और $B=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,जहाँ $x \in(0,1)$,तो $A-B=$

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