Question

Find the value of ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{3} \right)$

Answer 1

Hint: In this question, we have to find the value of ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{3} \right)$. We will first suppose both of the terms as separate variables and then use the property of tan(A+B) to simplify and evaluate the given terms. Formula which we will use is given by:
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
We will also use the basic value of $\tan \theta $ for finding our final answer.

Complete step by step answer:
Here we are given \[{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{3} \right)\]
To evaluate them easily, let us suppose value of ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)=A\text{ and }{{\tan }^{-1}}\left( \dfrac{1}{3} \right)=B$
Let us take tan on both sides in both terms, we get:
\[\tan \left( {{\tan }^{-1}}\dfrac{1}{2} \right)=\tan A\text{ and }\tan \left( {{\tan }^{-1}}\dfrac{1}{3} \right)=\tan B\]
Since tan and ${{\tan }^{-1}}$ are inverse terms, therefore they cancel out each other and we get:
\[\tan A=\dfrac{1}{2}\text{ and }\tan B=\dfrac{1}{3}\]
Now as we know,
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
Putting values of tanA and tanB in above equation, we get:
\[\tan \left( A+B \right)=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\left( \dfrac{1}{2} \right)\left( \dfrac{1}{3} \right)}\]
Now, let us take LCM on numerator we get:
\[\begin{align}
  & \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{3+2}{6}=\dfrac{5}{6} \\
 & \Rightarrow \tan \left( A+B \right)=\dfrac{\dfrac{5}{6}}{1-\dfrac{1}{6}} \\
\end{align}\]
Now, let us take LCM on denominator, we get:
\[\begin{align}
  & 1-\dfrac{1}{6}=\dfrac{6-1}{6}=\dfrac{5}{6} \\
 & \therefore \tan \left( A+B \right)=\dfrac{\dfrac{5}{6}}{\dfrac{5}{6}} \\
\end{align}\]
As we can see numerator and denominator are the same, so value becomes 1.
Hence, tan(A+B)=1.
Now, let us take ${{\tan }^{-1}}$ on both sides, we get:
\[\begin{align}
  & {{\tan }^{-1}}\tan \left( A+B \right)={{\tan }^{-1}}1 \\
 & \Rightarrow A+B={{\tan }^{-1}}1 \\
\end{align}\]
We have supposed earlier that ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)=A\text{ and }{{\tan }^{-1}}\left( \dfrac{1}{3} \right)=B$. Hence, putting those values in above, we get:
\[{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{3} \right)={{\tan }^{-1}}1\]
As we take $\tan \dfrac{\pi }{4}=1$. Therefore, taking ${{\tan }^{-1}}$ on both sides we get:
\[\begin{align}
  & {{\tan }^{-1}}\tan \dfrac{\pi }{4}={{\tan }^{-1}}1 \\
 & \therefore {{\tan }^{-1}}1=\dfrac{\pi }{4} \\
\end{align}\]
Putting this value in above expression, we get:
\[{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{3} \right)=\dfrac{\pi }{4}\]

Hence, value of \[{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{3} \right)=\dfrac{\pi }{4}\].

Note: Students should carefully perform calculations without getting confused between tan and ${{\tan }^{-1}}$. Students can directly learn the formula of ${{\tan }^{-1}}x+{{\tan }^{-1}}y$ which is given as
\[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]
Students should also learn basic values of $\tan \theta $ such as \[\tan {{0}^{4}}=0,\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}},\tan \dfrac{\pi }{4}=1\text{ and }\tan \dfrac{\pi }{3}=\sqrt{3}\]. Students should keep in mind that tan(A+B) is not just equal to tanA + tanB. But there exists proper formula for finding tan(A+B) which is
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]