sin^{-1} left( frac{12}{13} right) - sin^{-1} | vedclass

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(D) ધારો કે $\alpha = \sin^{-1} \left( \frac{12}{13} \right)$ અને $\beta = \sin^{-1} \left( \frac{3}{5} \right)$.તેથી $\sin \alpha = \frac{12}{13} \im

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$\frac{\pi}{2} - \sin^{-1} \left( \frac{56}{65} \right)$

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(D) ધારો કે $\alpha = \sin^{-1} \left( \frac{12}{13} ight)$ અને $\beta = \sin^{-1} \left( \frac{3}{5} ight)$.તેથી $\sin \alpha = \frac{12}{13} \implies \cos \alpha = \sqrt{1 - \left( \frac{12}{13} ight)^2} = \frac{5}{13}$.અને $\sin \beta = \frac{3}{5} \implies \cos \beta = \sqrt{1 - \left( \frac{3}{5} ight)^2} = \frac{4}{5}$.સૂત્ર $\sin^{-1} x - \sin^{-1} y = \sin^{-1} (x \sqrt{1 - y^2} - y \sqrt{1 - x^2})$ નો ઉપયોગ કરતા:$\sin^{-1} \left( \frac{12}{13} ight) - \sin^{-1} \left( \frac{3}{5} ight) = \sin^{-1} \left( \frac{12}{13} 	imes \frac{4}{5} - \frac{3}{5} 	imes \frac{5}{13} ight)$$= \sin^{-1} \left( \frac{48}{65} - \frac{15}{65} ight) = \sin^{-1} \left( \frac{33}{65} ight)$.આપણે જાણીએ છીએ કે $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,તેથી $\sin^{-1} \left( \frac{33}{65} ight) = \frac{\pi}{2} - \cos^{-1} \left( \frac{33}{65} ight)$.વૈકલ્પિક રીતે,$\cos^{-1} x = \sin^{-1} \sqrt{1 - x^2}$ નો ઉપયોગ કરતા,$\sin^{-1} \left( \frac{33}{65} ight) = \cos^{-1} \sqrt{1 - \left( \frac{33}{65} ight)^2} = \cos^{-1} \left( \frac{56}{65} ight) = \frac{\pi}{2} - \sin^{-1} \left( \frac{56}{65} ight)$.આમ,સાચો વિકલ્પ $D$ છે.

$\sin^{-1} \left( \frac{12}{13} \right) - \sin^{-1} \left( \frac{3}{5} \right)$ નું મૂલ્ય કેટલું થાય?

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