frac{tan 80^{circ}-tan 10^{circ}}{tan 70^{cir | vedclass

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(C) We know that $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Therefore,$	an A - 	an B = 	an(A-B)(1 + 	an A 	an B)$.Substituting $A =

Vedclass · Accepted Answer

$2$

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(C) We know that $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Therefore,$	an A - 	an B = 	an(A-B)(1 + 	an A 	an B)$.Substituting $A = 80^{\circ}$ and $B = 10^{\circ}$:$	an 80^{\circ} - 	an 10^{\circ} = 	an(80^{\circ}-10^{\circ})(1 + 	an 80^{\circ} 	an 10^{\circ}) = 	an 70^{\circ}(1 + 	an 80^{\circ} 	an 10^{\circ})$.Now,$\frac{	an 80^{\circ}-	an 10^{\circ}}{	an 70^{\circ}} = \frac{	an 70^{\circ}(1 + 	an 80^{\circ} 	an 10^{\circ})}{	an 70^{\circ}} = 1 + 	an 80^{\circ} 	an 10^{\circ}$.Since $	an 80^{\circ} = \cot 10^{\circ}$,we have $1 + \cot 10^{\circ} 	an 10^{\circ} = 1 + 1 = 2$.

$\frac{\tan 80^{\circ}-\tan 10^{\circ}}{\tan 70^{\circ}}$ is equal to

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