cos ^{-1}left{frac{1}{sqrt{2}}left(cos frac{9 | vedclass

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(C) हमारे पास $\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}ight)ight\}$ है।$\cos(\pi - 	heta) = -\cos 	h

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$\frac{17 \pi}{20}$

Vedclass · Answer

(C) हमारे पास $\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}ight)ight\}$ है।$\cos(\pi - 	heta) = -\cos 	heta$ और $\sin(\pi - 	heta) = \sin 	heta$ का उपयोग करने पर:$= \cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(-\cos \frac{\pi}{10} - \sin \frac{\pi}{10}ight)ight\}$$= \cos ^{-1}\left\{-\left(\frac{1}{\sqrt{2}} \cos \frac{\pi}{10} + \frac{1}{\sqrt{2}} \sin \frac{\pi}{10}ight)ight\}$$= \cos ^{-1}\left\{-\left(\cos \frac{\pi}{4} \cos \frac{\pi}{10} + \sin \frac{\pi}{4} \sin \frac{\pi}{10}ight)ight\}$$= \cos ^{-1}\left\{-\cos \left(\frac{\pi}{4} - \frac{\pi}{10}ight)ight\}$$= \cos ^{-1}\left\{-\cos \left(\frac{5\pi - 2\pi}{20}ight)ight\} = \cos ^{-1}\left\{-\cos \frac{3\pi}{20}ight\}$चूंकि $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$,इसलिए:$= \pi - \cos^{-1}\left(\cos \frac{3\pi}{20}ight)$$= \pi - \frac{3\pi}{20} = \frac{17\pi}{20}$.

$\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}$ का मान ज्ञात कीजिए।

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