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

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(B) આપેલ છે કે $\cos (\alpha + \beta ) = \frac{4}{5}$ અને $\sin (\alpha - \beta ) = \frac{5}{13}$.$0 < \alpha, \beta < \frac{\pi}{4}$ હોવાથી,$0 < \alp

Vedclass · Accepted Answer

$\frac{56}{33}$

Vedclass · Answer

(B) આપેલ છે કે $\cos (\alpha + \beta ) = \frac{4}{5}$ અને $\sin (\alpha - \beta ) = \frac{5}{13}$.$0 તેથી,$\sin (\alpha + \beta ) = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$ અને $\cos (\alpha - \beta ) = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$.આપણે જાણીએ છીએ કે $2\alpha = (\alpha + \beta ) + (\alpha - \beta )$.$\sin(A+B) = \sin A \cos B + \cos A \sin B$ સૂત્રનો ઉપયોગ કરતા:$\sin 2\alpha = \sin ((\alpha + \beta ) + (\alpha - \beta )) = \sin (\alpha + \beta ) \cos (\alpha - \beta ) + \cos (\alpha + \beta ) \sin (\alpha - \beta )$.$\sin 2\alpha = (\frac{3}{5} 	imes \frac{12}{13}) + (\frac{4}{5} 	imes \frac{5}{13}) = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}$.$\cos(A+B) = \cos A \cos B - \sin A \sin B$ સૂત્રનો ઉપયોગ કરતા:$\cos 2\alpha = \cos ((\alpha + \beta ) + (\alpha - \beta )) = \cos (\alpha + \beta ) \cos (\alpha - \beta ) - \sin (\alpha + \beta ) \sin (\alpha - \beta )$.$\cos 2\alpha = (\frac{4}{5} 	imes \frac{12}{13}) - (\frac{3}{5} 	imes \frac{5}{13}) = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$.તેથી,$	an 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{56/65}{33/65} = \frac{56}{33}$.

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

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