If tan 80^o = a and tan47^o = b, then tan37^o | vedclass

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Question

$	an 80^{\circ}=\alpha=\cot 10^{\circ}=\frac{1}{	an 10^{\circ}}$

$	an 47^{\circ}=\beta$

$	an 37^{\circ}=	an \left(47^{\circ}-10^{\circ}ig

Vedclass · Accepted Answer

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

Vedclass · Answer

$	an 80^{\circ}=\alpha=\cot 10^{\circ}=\frac{1}{	an 10^{\circ}}$

$	an 47^{\circ}=\beta$

$	an 37^{\circ}=	an \left(47^{\circ}-10^{\circ}ight)=\frac{	an 47^{\circ}-	an 10^{\circ}}{1+	an 47^{\circ} \cdot 	an 10^{\circ}}$

$=\frac{\beta-\frac{1}{\alpha}}{1+\frac{\beta}{\alpha}}=\frac{\alpha \beta-1}{\alpha+\beta}$

If $tan\ 80^o = a$ and $tan47^o = b$, then $tan37^o$ is equal to -

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