જો cos left( frac{alpha - beta}{2} right) = 2 | vedclass

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(B) આપેલ સમીકરણ: $\cos \left( \frac{\alpha - \beta}{2} \right) = 2\cos \left( \frac{\alpha + \beta}{2} \right)$.વિસ્તરણ સૂત્રો $\cos(A - B) = \cos A \

Vedclass · Accepted Answer

$\frac{1}{3}$

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(B) આપેલ સમીકરણ: $\cos \left( \frac{\alpha - \beta}{2} ight) = 2\cos \left( \frac{\alpha + \beta}{2} ight)$.વિસ્તરણ સૂત્રો $\cos(A - B) = \cos A \cos B + \sin A \sin B$ અને $\cos(A + B) = \cos A \cos B - \sin A \sin B$ નો ઉપયોગ કરતા:$\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \left( \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - \sin \frac{\alpha}{2} \sin \frac{\beta}{2} ight)$.જમણી બાજુનું વિસ્તરણ કરતા:$\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - 2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}$.પદોને ગોઠવતા:$3 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$.બંને બાજુ $3 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$ વડે ભાગતા,આપણને મળે:$\frac{\sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{\cos \frac{\alpha}{2} \cos \frac{\beta}{2}} = \frac{1}{3}$.તેથી,$	an \frac{\alpha}{2} 	an \frac{\beta}{2} = \frac{1}{3}$.

જો $\cos \left( \frac{\alpha - \beta}{2} \right) = 2\cos \left( \frac{\alpha + \beta}{2} \right)$ આપેલ હોય,તો $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}$ ની કિંમત શોધો.

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