જો m cos (alpha+beta)-n cos (alpha-beta)=m co | vedclass

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(C) આપેલ છે: $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$પદોને ગોઠવતા:$m [\cos (\alpha+\beta) - \cos (\al

Vedclass · Accepted Answer

$-\frac{n}{m}$

Vedclass · Answer

(C) આપેલ છે: $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$પદોને ગોઠવતા:$m [\cos (\alpha+\beta) - \cos (\alpha-\beta)] = n [\cos (\alpha+\beta) + \cos (\alpha-\beta)]$નિત્યસમ $\cos (A+B) - \cos (A-B) = -2 \sin A \sin B$ અને $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$ નો ઉપયોગ કરતા:$m [-2 \sin \alpha \sin \beta] = n [2 \cos \alpha \cos \beta]$$-2m \sin \alpha \sin \beta = 2n \cos \alpha \cos \beta$બંને બાજુ $2m \cos \alpha \cos \beta$ વડે ભાગતા:$\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = -\frac{n}{m}$$	an \alpha 	an \beta = -\frac{n}{m}$

જો $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$ હોય,તો $\tan \alpha \tan \beta=$

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