sin left[ cos^{-1} left( frac{3}{5} right) + | vedclass

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Question

(A) माना $\alpha = \cos^{-1} \left( \frac{3}{5} \right)$,तो $\cos \alpha = \frac{3}{5}$। चूँकि $\cos^2 \alpha + \sin^2 \alpha = 1$,इसलिए $\sin \alpha

Vedclass · Accepted Answer

$\frac{2}{\sqrt{5}}$

Vedclass · Answer

(A) माना $\alpha = \cos^{-1} \left( \frac{3}{5} ight)$,तो $\cos \alpha = \frac{3}{5}$। चूँकि $\cos^2 \alpha + \sin^2 \alpha = 1$,इसलिए $\sin \alpha = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$। अतः,$\alpha = \sin^{-1} \left( \frac{4}{5} ight)$।माना $\beta = 	an^{-1} 2$,तो $	an \beta = 2$। सर्वसमिका $\sin \beta = \frac{	an \beta}{\sqrt{1 + 	an^2 \beta}}$ का उपयोग करने पर,$\sin \beta = \frac{2}{\sqrt{1 + 2^2}} = \frac{2}{\sqrt{5}}$ प्राप्त होता है। अतः,$\beta = \sin^{-1} \left( \frac{2}{\sqrt{5}} ight)$।व्यंजक $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ बन जाता है।यहाँ $\sin \alpha = \frac{4}{5}$,$\cos \alpha = \frac{3}{5}$,$\sin \beta = \frac{2}{\sqrt{5}}$,और $\cos \beta = \frac{1}{\sqrt{1 + 2^2}} = \frac{1}{\sqrt{5}}$ है।इन मानों को प्रतिस्थापित करने पर: $\sin(\alpha + \beta) = (\frac{4}{5} 	imes \frac{1}{\sqrt{5}}) + (\frac{3}{5} 	imes \frac{2}{\sqrt{5}}) = \frac{4}{5\sqrt{5}} + \frac{6}{5\sqrt{5}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$।

$\sin \left[ \cos^{-1} \left( \frac{3}{5} \right) + \tan^{-1} 2 \right] = $

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$\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots$ अनंत तक का मान ज्ञात कीजिए।

यदि $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ है,जहाँ $0 < x < 1$,तो $\frac{dy}{dx}$ ज्ञात कीजिए।

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