Question

Given that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ where A & B are acute angles. Calculate A + B when $\tan A=\dfrac{1}{2},\tan B=\dfrac{1}{3}$.
(a) A + B = 30°
(b) A +B = 45°
(c) A + B =60°
(d) A +B = 75°

Answer 1

Hint: First of all substitute the values of tan A and tan B as mentioned in the question and then solve the right hand expression of the equation. After solving the right hand side expression, we will come with an expression like tan (A + B) = 1 then A + B is the angle where tan becomes 1.

Complete step-by-step answer:
We are solving the right hand side of the expression in tan given in the question by substituting the values of tan A and tan B as mentioned in the question as follows:
$\begin{align}
  & \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \\
 & \Rightarrow \tan \left( A+B \right)=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\dfrac{1}{2}.\dfrac{1}{3}} \\
 & \Rightarrow \tan \left( A+B \right)=\dfrac{\dfrac{3+2}{6}}{\dfrac{6-1}{6}} \\
 & \Rightarrow \tan \left( A+B \right)=\dfrac{\dfrac{5}{6}}{\dfrac{5}{6}} \\
\end{align}$
$\Rightarrow \tan \left( A+B \right)=1$
We know that if tan θ = 1 then θ = 45° so we can use this relation in the above expression we get,
A + B = 45°
Hence, the correct option is (b).

Note: In the above solution, there is a step in which we are getting tan (A + B) =1. Now, it has many solutions because tan is positive in the first and third quadrants but it is given in the question that A & B are acute angles so we have to take the angle which lies in the first quadrant. We can see from the options also that all the angles are in the first quadrant so if we have ignored that A and B are acute angles still we will get the correct answer.