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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 = $

Similar Questions

$\cos(18^{\circ}-A) \cos(18^{\circ}+A) - \cos(72^{\circ}-A) \cos(72^{\circ}+A)$ ની કિંમત શોધો.

જો બે ખૂણાઓ $\alpha, \beta$ એવા હોય કે $0 < \alpha, \beta < \frac{\pi}{4}$,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ અને $\frac{\sqrt{1-\cos 2 \beta}}{\sqrt{1+\cos 2 \beta}}=\frac{1}{7}$,તો $(2 \alpha+\beta)=$

$\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos ^{2} 30^{\circ}-\cos ^{2} 60^{\circ}$ નું મૂલ્ય શોધો.

$\frac{\sin(B + A) + \cos(B - A)}{\sin(B - A) + \cos(B + A)} = $

જો $\cos (A - B) = \frac{3}{5}$ અને $\tan A \tan B = 2$ હોય,તો

Vedclass Test Series

Exam Paper Generator

Online Exam Module