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

જો $y = \sin^{-1}(2x\sqrt{1-x^2})$ હોય અને $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$ હોય,તો $\frac{dy}{dx}$ શોધો.

જો $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$,જ્યાં $-1 \le x \le 1$ હોય,તો $\frac{dy}{dx}$ શોધો.

$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}$ નું મૂલ્ય શોધો.

જો $(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$ હોય,તો $x$ ની કિંમત શોધો.

જો $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$ હોય,તો $p^2 + q^2 + r^2 + 2pqr = $

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