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: Apply the given formula for the tangent of sum of two angles given by: - \[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\] and substitute the values of \[\tan A\] and \[\tan B\] given the question. If the R.H.S value, after simplifying the expression is 1 then take \[{{\tan }^{-1}}\] function both side and use the information \[{{\tan }^{-1}}1={{45}^{\circ }}\].

Complete step-by-step solution
We have been provided with the formula for the tangent of sum of two angles, which is: -
\[\Rightarrow \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
Also, we have been provided with the information: - \[\tan A=\dfrac{1}{3}\] and \[\tan B=\dfrac{1}{2}\].
So, let us find the value of \[\tan \left( A+B \right)\] by substituting the values of \[\tan A\] and \[\tan B\] at their respective position in the above formula.
\[\begin{align}
  & \Rightarrow \tan \left( A+B \right)=\dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{1-\dfrac{1}{3}\times \dfrac{1}{2}} \\
 & \Rightarrow \tan \left( A+B \right)=\dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{1-\dfrac{1}{6}} \\
\end{align}\]
Taking L.C.M in the numerator and denominator, we have,
\[\Rightarrow \tan \left( A+B \right)=\dfrac{\dfrac{2+3}{6}}{\dfrac{6-1}{6}}\]
\[\begin{align}
  & \Rightarrow \tan \left( A+B \right)=\dfrac{5}{5} \\
 & \Rightarrow \tan \left( A+B \right)=1 \\
\end{align}\]
Taking inverse function both sides, we get,
\[\Rightarrow {{\tan }^{-1}}\left[ \tan \left( A+B \right) \right]={{\tan }^{-1}}1\]
Using the identity \[{{\tan }^{-1}}\left( \tan x \right)=x\], we have,
\[\Rightarrow A+B={{\tan }^{-1}}1\] - (i)
Now, we know that, A and B are acute angles therefore, we have to consider A and B from the \[{{1}^{st}}\] quadrant,
\[\begin{align}
  & \Rightarrow \tan {{45}^{\circ }}=1 \\
 & \Rightarrow {{\tan }^{-1}}1={{45}^{\circ }} \\
\end{align}\]
Therefore, from equation (i), we have,
\[\Rightarrow A+B={{45}^{\circ }}\]
Hence, proved

Note: One may note that we don’t have any other option but to use the given formula. Here, we don’t know the values of angle A and B for which their tangent values are \[\dfrac{1}{3}\] and \[\dfrac{1}{2}\] respectively. These tangent values don’t have an angle which we use to remember like \[{{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }}\], etc. So, it is important to use the given formula. One of the most important things to note is that there are many values of angles for which its tangent value is \[{{45}^{\circ }}\] but since we have been provided that A and B are acute and so is A + B, therefore we have to take the value of A + B from the first quadrant.