Question

If A and B are acute angles such that $\tan A=\dfrac{1}{2}$ ,$\tan B=\dfrac{1}{3}$ and $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ , find A+B

Answer 1

Hint:The value of tanA and tanB is given, so we will substitute that value in $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ and find the value tan(A+B) and then it is given that A and B are acute angles so with the help of that we will find the value of A + B, such that the condition is satisfied.


Complete step-by-step answer:
It is given that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$,
Now we will substitute $\tan A=\dfrac{1}{2}$ ,$\tan B=\dfrac{1}{3}$ in $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
Therefore, after substituting we get,
$\begin{align}
  & \tan \left( A+B \right)=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\dfrac{1}{2}\times \dfrac{1}{3}} \\
 & \tan \left( A+B \right)=\dfrac{\dfrac{5}{6}}{\dfrac{5}{6}} \\
 & \tan \left( A+B \right)=1 \\
\end{align}$
Now we know that tan45 = 1, and tan is positive in the 1st and 3rd quadrant but the value of A + B in 3rd quadrant will not give any value of A and B such that both of them are acute.
Hence, the only option remaining for us is A + B = 45.
Hence, we have found the desired value.

Note: One important point one should keep in mind is that there are infinitely many angles for which we get the value of tan as 1, and many of the students get stuck at this point and then make mistakes. But we have used a small but very important point that is given in the question is that both A and B must be acute angles which helps us to find one correct answer from the infinitely many possibilities.