If tan 80^{circ} = alpha and tan 47^{circ} = | vedclass

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(C) Given $	an 80^{\circ} = \alpha$. Since $	an 80^{\circ} = \cot 10^{\circ}$,we have $\cot 10^{\circ} = \alpha$,which implies $	an 10^{\circ} = \f

Vedclass · Accepted Answer

$\frac{\alpha \beta - 1}{\alpha + \beta}$

Vedclass · Answer

(C) Given $	an 80^{\circ} = \alpha$. Since $	an 80^{\circ} = \cot 10^{\circ}$,we have $\cot 10^{\circ} = \alpha$,which implies $	an 10^{\circ} = \frac{1}{\alpha}$.Given $	an 47^{\circ} = \beta$.We need to find $	an 37^{\circ}$. We can write $37^{\circ} = 47^{\circ} - 10^{\circ}$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$:$	an 37^{\circ} = 	an(47^{\circ} - 10^{\circ}) = \frac{	an 47^{\circ} - 	an 10^{\circ}}{1 + 	an 47^{\circ} 	an 10^{\circ}}$Substituting the values:$= \frac{\beta - \frac{1}{\alpha}}{1 + \beta \cdot \frac{1}{\alpha}} = \frac{\frac{\alpha \beta - 1}{\alpha}}{\frac{\alpha + \beta}{\alpha}} = \frac{\alpha \beta - 1}{\alpha + \beta}$

If $\tan 80^{\circ} = \alpha$ and $\tan 47^{\circ} = \beta$,then $\tan 37^{\circ}$ is equal to -

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