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

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Question

(B) ધારો કે $x = \frac{\sqrt{3}}{2}$ અને $y = \sqrt{\frac{2}{3}}$.નોંધો કે $x^2 + y^2 = \frac{3}{4} + \frac{2}{3} = \frac{9+8}{12} = \frac{17}{12} > 1

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) ધારો કે $x = \frac{\sqrt{3}}{2}$ અને $y = \sqrt{\frac{2}{3}}$.નોંધો કે $x^2 + y^2 = \frac{3}{4} + \frac{2}{3} = \frac{9+8}{12} = \frac{17}{12} > 1$.જ્યારે $x, y > 0$ અને $x^2 + y^2 > 1$ હોય,ત્યારે આપણે નિત્યસમ $\sin^{-1} x + \sin^{-1} y = \pi - \sin^{-1} (x \sqrt{1-y^2} + y \sqrt{1-x^2})$ નો ઉપયોગ કરીએ છીએ.કિંમતો મૂકતા:$\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

$\tan \left[ 2\tan^{-1}\left( \frac{1}{5} \right) - \frac{\pi}{4} \right] = $

સમીકરણ $\sin^{-1} x = 2\tan^{-1} x$ નો ઉકેલ ગણ શોધો.

$\sin^{-1} \frac{1}{\sqrt{5}} + \cot^{-1} 3$ ની કિંમત શોધો.

જો $\frac{1}{2} \sin^{-1}\left(\frac{3 \sin 2\theta}{5+4 \cos 2\theta}\right) = \tan^{-1} x$ હોય,તો $x =$

$\tan \left(2 \tan ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)\right)$ ની કિંમત શોધો:

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