tan 100^circ + tan 125^circ + tan 100^circ ta | vedclass

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(D) We use the trigonometric identity: $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Let $A = 100^\circ$ and $B = 125^\circ$.Then,$	an(10

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We use the trigonometric identity: $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Let $A = 100^\circ$ and $B = 125^\circ$.Then,$	an(100^\circ + 125^\circ) = \frac{	an 100^\circ + 	an 125^\circ}{1 - 	an 100^\circ 	an 125^\circ}$.Since $100^\circ + 125^\circ = 225^\circ$,we have $	an 225^\circ = 	an(180^\circ + 45^\circ) = 	an 45^\circ = 1$.Substituting this into the identity:$1 = \frac{	an 100^\circ + 	an 125^\circ}{1 - 	an 100^\circ 	an 125^\circ}$.Multiplying both sides by $(1 - 	an 100^\circ 	an 125^\circ)$:$1 - 	an 100^\circ 	an 125^\circ = 	an 100^\circ + 	an 125^\circ$.Rearranging the terms:$	an 100^\circ + 	an 125^\circ + 	an 100^\circ 	an 125^\circ = 1$.

$\tan 100^\circ + \tan 125^\circ + \tan 100^\circ \tan 125^\circ = $

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