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

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(B) આપેલ છે કે $\cos(\alpha + \beta) = \frac{4}{5}$. કારણ કે $0 \le \alpha, \beta \le \frac{\pi}{4}$,તેથી $0 \le \alpha + \beta \le \frac{\pi}{2}$,માટ

Vedclass · Accepted Answer

$\frac{56}{33}$

Vedclass · Answer

(B) આપેલ છે કે $\cos(\alpha + \beta) = \frac{4}{5}$. કારણ કે $0 \le \alpha, \beta \le \frac{\pi}{4}$,તેથી $0 \le \alpha + \beta \le \frac{\pi}{2}$,માટે $	an(\alpha + \beta) = \frac{3}{4}$.આપેલ છે કે $\sin(\alpha - \beta) = \frac{5}{13}$. કારણ કે $0 \le \alpha, \beta \le \frac{\pi}{4}$,તેથી $-\frac{\pi}{4} \le \alpha - \beta \le \frac{\pi}{4}$,માટે $	an(\alpha - \beta) = \frac{5}{12}$.આપણે જાણીએ છીએ કે $2\alpha = (\alpha + \beta) + (\alpha - \beta)$.સૂત્ર $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ નો ઉપયોગ કરતા:$	an 2\alpha = 	an((\alpha + \beta) + (\alpha - \beta)) = \frac{	an(\alpha + \beta) + 	an(\alpha - \beta)}{1 - 	an(\alpha + \beta)	an(\alpha - \beta)}$$	an 2\alpha = \frac{\frac{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4} 	imes \frac{5}{12})} = \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{14}{12} 	imes \frac{48}{33} = \frac{14 	imes 4}{33} = \frac{56}{33}$.

જો $\cos(\alpha + \beta) = \frac{4}{5}$ અને $\sin(\alpha - \beta) = \frac{5}{13}$,જ્યાં $0 \le \alpha, \beta \le \frac{\pi}{4}$ હોય,તો $\tan 2\alpha = $

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