Question

find the exact value of the expression.
\\( \frac { \tan 50 ^ { \circ } + \tan 130 ^ { \circ } } { 1 - \tan 50 ^ { \circ } \tan 130 ^ { \circ } } \\)
rewrite the expression using a sum or difference formula. choose the correct ans
\\( \frac { \tan 50 ^ { \circ } + \tan 130 ^ { \circ } } { 1 - \tan 50 ^ { \circ } \tan 130 ^ { \circ } } = \tan 130 ^ { \circ } - \tan 50 ^ { \circ } = \tan 80 ^ { \circ } \\)
\\( \frac { \tan 50 ^ { \circ } + \tan 130 ^ { \circ } } { 1 - \tan 50 ^ { \circ } \tan 130 ^ { \circ } } = \tan \left( 130 ^ { \circ } + 50 ^ { \circ } \
ight) = \tan 180 ^ { \circ } \\)
\\( \frac { \tan 50 ^ { \circ } + \tan 130 ^ { \circ } } { 1 - \tan 50 ^ { \circ } \tan 130 ^ { \circ } } = \tan \left( 130 ^ { \circ } - 50 ^ { \circ } \
ight) = \tan 80 ^ { \circ } \\)
\\( \frac { \tan 50 ^ { \circ } + \tan 130 ^ { \circ } } { 1 - \tan 50 ^ { \circ } \tan 130 ^ { \circ } } = \tan 130 ^ { \circ } + \tan 50 ^ { \circ } = \tan 180 ^ { \circ } \\)
\\( \frac { \tan 50 ^ { \circ } + \tan 130 ^ { \circ } } { 1 - \tan 50 ^ { \circ } \tan 130 ^ { \circ } } = \square \\)
(simplify your answer, including any radicals. use integers or fractions for any n

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:

$0$