Question

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

Answer 1

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:

& \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{aligned}$$ Now, let us take LCM on denominator, we get: $$\begin{aligned} & 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{aligned}$$ As we can see numerator and denominator are the same, so value becomes 1. Hence, tan(A+B)=1. Now, let us take ${{\tan }^{-1}}$ on both sides, we get: $$\begin{aligned} & {{\tan }^{-1}}\tan \left( A+B \right)={{\tan }^{-1}}1 \\\ & \Rightarrow A+B={{\tan }^{-1}}1 \\\ \end{aligned}$$ We have supposed earlier that ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)=A\text{ and }{{\tan }^{-1}}\left( \dfrac{1}{3} \right)=B$. Hence, putting those values in above, we get: $${{\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{aligned} & {{\tan }^{-1}}\tan \dfrac{\pi }{4}={{\tan }^{-1}}1 \\\ & \therefore {{\tan }^{-1}}1=\dfrac{\pi }{4} \\\ \end{aligned}$$ Putting this value in above expression, we get: $${{\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}$$