यदि cos(alpha + beta) = frac{3}{5},sin(alpha | vedclass

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(C) दिया गया है कि $0 < \alpha, \beta < \frac{\pi}{4}$,इसलिए $0 < \alpha + \beta < \frac{\pi}{2}$ और $-\frac{\pi}{4} < \alpha - \beta < \frac{\pi}{4}$

Vedclass · Accepted Answer

$\frac{63}{16}$

Vedclass · Answer

(C) दिया गया है कि $0 चूंकि $\cos(\alpha + \beta) = \frac{3}{5}$,इसलिए $	an(\alpha + \beta) = \frac{\sqrt{1 - (3/5)^2}}{3/5} = \frac{4/5}{3/5} = \frac{4}{3}$ होगा।चूंकि $\sin(\alpha - \beta) = \frac{5}{13}$,इसलिए $	an(\alpha - \beta) = \frac{5/13}{\sqrt{1 - (5/13)^2}} = \frac{5/13}{12/13} = \frac{5}{12}$ होगा।अब,$	an(2\alpha) = 	an((\alpha + \beta) + (\alpha - \beta))$ होगा।सूत्र $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ का उपयोग करने पर:$	an(2\alpha) = \frac{\frac{4}{3} + \frac{5}{12}}{1 - (\frac{4}{3} \cdot \frac{5}{12})} = \frac{\frac{16+5}{12}}{1 - \frac{20}{36}} = \frac{21/12}{16/36} = \frac{21}{12} \cdot \frac{36}{16} = \frac{21 \cdot 3}{16} = \frac{63}{16}$।

यदि $\cos(\alpha + \beta) = \frac{3}{5}$,$\sin(\alpha - \beta) = \frac{5}{13}$ और $0 < \alpha, \beta < \frac{\pi}{4}$ है,तो $\tan(2\alpha)$ का मान ज्ञात कीजिए।

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