Question

If A and B are acute angles such that \[\tan A=\dfrac{1}{3},\tan B=\dfrac{1}{2}\] and $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. Show that $A+B=45{}^\circ $.

Answer 1

Hint: We will be using the concept of trigonometric functions to solve the problem. We will use the value of \[\tan A=\dfrac{1}{3}\ and\ \tan B=\dfrac{1}{2}\] to simplify the RHS and then use it to prove that $A+B=45{}^\circ $.

Complete step-by-step answer:


Now, we have been given that A and B are acute angles such that \[\tan A=\dfrac{1}{3},\tan B=\dfrac{1}{2}\] and $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}..........\left( 1 \right)$

We have to prove that $A+B=45{}^\circ $.

Now, we will first take RHS of equation (1) and substitute the value of $\tan A,\tan B$ in it. So, in LHS we have that,

\[\begin{align}

  & \dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{1-\dfrac{1}{3}\times \dfrac{1}{2}} \\

 & =\dfrac{\dfrac{2+3}{6}}{1-\dfrac{1}{6}} \\

 & =\dfrac{\dfrac{5}{6}}{\dfrac{6-1}{6}} \\

 & =\dfrac{\dfrac{5}{6}}{\dfrac{5}{6}} \\

\end{align}\]

Since numerator and denominator are same, therefore the value of RHS is,

$\dfrac{\tan A+\tan B}{1-\tan A\tan B}=1$

Now, we use this in equation (1). So, we have,

$\tan \left( A+B \right)=1$

Now, we know that $1=\tan 45{}^\circ $. So,

$\tan \left( A+B \right)=\tan 45{}^\circ $

On comparing both we have,

$A+B=45{}^\circ $

Hence, proved.


Note: To solve these types of questions it is important to note that we have substituted the values of $\tan A\ and\ \tan B$in RHS and used its value to prove that $A+B=45{}^\circ $.