find the exact value of the expression.\n\\( \\frac { \\tan 50 ^ { \\circ } + \\tan 130 ^ { \\circ } } { 1…

find the exact value of the expression.\n\\( \\frac { \\tan 50 ^ { \\circ } + \\tan 130 ^ { \\circ } } { 1 - \\tan 50 ^ { \\circ } \\tan 130 ^ { \\circ } } \\)\nrewrite the expression using a sum or difference formula. choose the correct ans\n\\( \\frac { \\tan 50 ^ { \\circ } + \\tan 130 ^ { \\circ } } { 1 - \\tan 50 ^ { \\circ } \\tan 130 ^ { \\circ } } = \\tan 130 ^ { \\circ } - \\tan 50 ^ { \\circ } = \\tan 80 ^ { \\circ } \\)\n\\( \\frac { \\tan 50 ^ { \\circ } + \\tan 130 ^ { \\circ } } { 1 - \\tan 50 ^ { \\circ } \\tan 130 ^ { \\circ } } = \\tan \\left( 130 ^ { \\circ } + 50 ^ { \\circ } \\right) = \\tan 180 ^ { \\circ } \\)\n\\( \\frac { \\tan 50 ^ { \\circ } + \\tan 130 ^ { \\circ } } { 1 - \\tan 50 ^ { \\circ } \\tan 130 ^ { \\circ } } = \\tan \\left( 130 ^ { \\circ } - 50 ^ { \\circ } \\right) = \\tan 80 ^ { \\circ } \\)\n\\( \\frac { \\tan 50 ^ { \\circ } + \\tan 130 ^ { \\circ } } { 1 - \\tan 50 ^ { \\circ } \\tan 130 ^ { \\circ } } = \\tan 130 ^ { \\circ } + \\tan 50 ^ { \\circ } = \\tan 180 ^ { \\circ } \\)\n\\( \\frac { \\tan 50 ^ { \\circ } + \\tan 130 ^ { \\circ } } { 1 - \\tan 50 ^ { \\circ } \\tan 130 ^ { \\circ } } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any n
Answer
Explanation:
Step1: Recall the tangent sum formula
The tangent sum formula is $\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. Let $A = 50^{\circ}$ and $B=130^{\circ}$.
Step2: Apply the formula
Substituting $A = 50^{\circ}$ and $B = 130^{\circ}$ into the formula $\frac{\tan A+\tan B}{1-\tan A\tan B}$, we get $\frac{\tan50^{\circ}+\tan130^{\circ}}{1 - \tan50^{\circ}\tan130^{\circ}}=\tan(50^{\circ}+130^{\circ})$.
Step3: Calculate the angle
$50^{\circ}+130^{\circ}=180^{\circ}$.
Step4: Find the value of $\tan180^{\circ}$
We know that $\tan180^{\circ}=0$.
Answer:
$0$