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 answer below.
\\( \bigcirc \\) a. \\( \frac { \tan 50 ^ { \circ } + \tan 130 ^ { \circ } } { 1 - \tan 50 ^ { \circ } \tan 130 ^ { \circ } } = \tan 130 ^ { \circ } - \tan 50 ^ { \circ } = \tan 80 ^ { \circ } \\)
\\( \bigcirc \\) b. \\( \frac { \tan 50 ^ { \circ } + \tan 130 ^ { \circ } } { 1 - \tan 50 ^ { \circ } \tan 130 ^ { \circ } } = \tan ( 130 ^ { \circ } + 50 ^ { \circ } ) = \tan 180 ^ { \circ } \\)
\\( \bigcirc \\) c. \\( \frac { \tan 50 ^ { \circ } + \tan 130 ^ { \circ } } { 1 - \tan 50 ^ { \circ } \tan 130 ^ { \circ } } = \tan ( 130 ^ { \circ } - 50 ^ { \circ } ) = \tan 80 ^ { \circ } \\)
\\( \bigcirc \\) d. \\( \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 \\)

Explanation:

Step1: Recall the tangent subtraction formula

The formula for \(\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}\), and its variation \(\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A + B)\)

Step2: Identify \(A\) and \(B\)

Let \(A = 130^{\circ}\) and \(B=50^{\circ}\)

Step3: Apply the formula

\(\frac{\tan50^{\circ}+\tan130^{\circ}}{1 - \tan50^{\circ}\tan130^{\circ}}=\tan(50^{\circ}+ 130^{\circ})\)

Step4: Calculate \(A + B\)

\(50^{\circ}+130^{\circ}=180^{\circ}\)

Answer:

B. \(\frac{\tan50^{\circ}+\tan130^{\circ}}{1-\tan50^{\circ}\tan130^{\circ}}=\tan(130^{\circ}+50^{\circ})=\tan180^{\circ}\)