Question 36

If angles $A$ and $B$ satisfy the system of equations

2 \tan A + \tan B = 4

\tan A - 3 \tan B = -\frac{17}{2}

Then $\tan(2A + B)$ equals....

Explanation

Recall the trigonometric concepts for angle addition

\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \cdot \tan y}

\tan(2x) = \frac{2 \tan x}{1 - \tan^2 x}

Solve the system of equations

Get the first equation by transforming to find $\tan B$ .

2 \tan A + \tan B = 4 \rightarrow \tan B = 4 - 2 \tan A

Then substitute into the second equation

\tan A - 3 \tan B = -\frac{17}{2}

\tan A - 3(4 - 2 \tan A) = -\frac{17}{2}

\tan A - 12 + 6 \tan A = -\frac{17}{2}

7 \tan A = -\frac{17}{2} + 12

7 \tan A = \frac{7}{2}

\tan A = \frac{1}{2}

Therefore, the value of $\tan B = 4 - 2 \tan A = 4 - 2\left(\frac{1}{2}\right) = 3$ .

Determine the value of tan 2A

\tan(2A) = \frac{2 \tan A}{1 - \tan^2 A}

\tan(2A) = \frac{2\left(\frac{1}{2}\right)}{1 - \left(\frac{1}{2}\right)^2}

\tan(2A) = \frac{1}{1 - \frac{1}{4}} = \frac{4}{3}

Determine the value of tan(2A + B)

\tan(2A + B) = \frac{\tan 2A + \tan B}{1 - \tan 2A \cdot \tan B}

\tan(2A + B) = \frac{\frac{4}{3} + 3}{1 - \frac{4}{3}(3)}

\tan(2A + B) = \frac{\frac{13}{3}}{1 - 4}

\tan(2A + B) = \frac{\frac{13}{3}}{-3}

\tan(2A + B) = -\frac{13}{9}

Therefore, the value of $\tan(2A + B)$ is $-\frac{13}{9}$ .