The value of tan 40^{circ} + tan 11^{circ} + | vedclass

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Question

(A) We know that $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Consider the identity $	an 60^{\circ} = 	an(40^{\circ} + 20^{\circ}) = \fra

Vedclass · Accepted Answer

$\sqrt{3}-1$

Vedclass · Answer

(A) We know that $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Consider the identity $	an 60^{\circ} = 	an(40^{\circ} + 20^{\circ}) = \frac{	an 40^{\circ} + 	an 20^{\circ}}{1 - 	an 40^{\circ} 	an 20^{\circ}} = \sqrt{3}$.This implies $	an 40^{\circ} + 	an 20^{\circ} = \sqrt{3}(1 - 	an 40^{\circ} 	an 20^{\circ}) = \sqrt{3} - \sqrt{3} 	an 40^{\circ} 	an 20^{\circ}$.Rearranging gives $	an 40^{\circ} + 	an 20^{\circ} + \sqrt{3} 	an 40^{\circ} 	an 20^{\circ} = \sqrt{3}$.Now consider $	an(56^{\circ} - 11^{\circ}) = 	an 45^{\circ} = 1$.Using the formula $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we have $\frac{	an 56^{\circ} - 	an 11^{\circ}}{1 + 	an 56^{\circ} 	an 11^{\circ}} = 1$.This implies $	an 56^{\circ} - 	an 11^{\circ} = 1 + 	an 56^{\circ} 	an 11^{\circ}$,or $	an 56^{\circ} - 	an 11^{\circ} - 	an 56^{\circ} 	an 11^{\circ} = 1$.Substituting these into the original expression: $(	an 40^{\circ} + 	an 20^{\circ} + \sqrt{3} 	an 40^{\circ} 	an 20^{\circ}) + (	an 11^{\circ} - 	an 56^{\circ} + 	an 56^{\circ} 	an 11^{\circ})$.From the second identity,$	an 56^{\circ} - 	an 11^{\circ} - 	an 56^{\circ} 	an 11^{\circ} = 1$,so $	an 11^{\circ} - 	an 56^{\circ} + 	an 56^{\circ} 	an 11^{\circ} = -1$.Thus,the total value is $\sqrt{3} - 1$.

The value of $\tan 40^{\circ} + \tan 11^{\circ} + \tan 20^{\circ} - \tan 56^{\circ} + \tan 56^{\circ} \tan 11^{\circ} + \sqrt{3} \tan 40^{\circ} \tan 20^{\circ}$ is

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