यदि cos(alpha + beta) = frac{4}{5} और sin(alp | vedclass

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(B) दिया है $\cos(\alpha + \beta) = \frac{4}{5}$. चूँकि $0 \le \alpha, \beta \le \frac{\pi}{4}$,इसलिए $0 \le \alpha + \beta \le \frac{\pi}{2}$,अतः $

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$\frac{56}{33}$

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(B) दिया है $\cos(\alpha + \beta) = \frac{4}{5}$. चूँकि $0 \le \alpha, \beta \le \frac{\pi}{4}$,इसलिए $0 \le \alpha + \beta \le \frac{\pi}{2}$,अतः $	an(\alpha + \beta) = \frac{3}{4}$.दिया है $\sin(\alpha - \beta) = \frac{5}{13}$. अतः $	an(\alpha - \beta) = \frac{5}{12}$.अब,$	an 2\alpha = 	an((\alpha + \beta) + (\alpha - \beta))$.$	an 2\alpha = \frac{\frac{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4} 	imes \frac{5}{12})} = \frac{14/12}{33/48} = \frac{56}{33}$.

यदि $\cos(\alpha + \beta) = \frac{4}{5}$ और $\sin(\alpha - \beta) = \frac{5}{13}$,जहाँ $0 \le \alpha, \beta \le \frac{\pi}{4}$ है,तो $\tan 2\alpha = $

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