Question

How do you find the value of $ {{\tan }^{-1}}\dfrac{2}{3} $ ?

Answer 1

Hint: If the value of tan x is equal to y where x is in between $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ then we can write $ x={{\tan }^{-1}}y $ , so if we have to find the value of $ {{\tan }^{-1}}\dfrac{2}{3} $ then we have find a x in the range $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ such that $ \tan x=\dfrac{2}{3} $

Complete step by step answer:
We have find the value of $ {{\tan }^{-1}}\dfrac{2}{3} $ so we have to find a value x in between $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ such that $ \tan x=\dfrac{2}{3} $
We can see that there is no standard angle such as 0, $ \dfrac{\pi }{6} $ , $ \dfrac{\pi }{3} $ , $ \dfrac{\pi }{12} $ etc. such that $ \tan x=\dfrac{2}{3} $
So we can not find the value of $ {{\tan }^{-1}}\dfrac{2}{3} $ manually, but we can transform $ {{\tan }^{-1}}\dfrac{2}{3} $ into other form
We know the formula $ {{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right) $ where xy is not equal to -1
We can write $ \dfrac{2}{3} $ as $ \dfrac{1-\dfrac{1}{5}}{1+1\times \dfrac{1}{5}} $
We can write $ {{\tan }^{-1}}\dfrac{2}{3} $ = $ {{\tan }^{-1}}\dfrac{1-\dfrac{1}{5}}{1+1\times \dfrac{1}{5}} $
Taking x as 1 and y as $ \dfrac{1}{5} $ , and applying the formula we get
 $ {{\tan }^{-1}}\dfrac{1-\dfrac{1}{5}}{1+1\times \dfrac{1}{5}}={{\tan }^{-1}}1-{{\tan }^{-1}}\dfrac{1}{5} $
We know the value of $ {{\tan }^{-1}}1 $ is equal to $ \dfrac{\pi }{4} $
So we can write $ {{\tan }^{-1}}\dfrac{2}{3}=\dfrac{\pi }{4}-{{\tan }^{-1}}\dfrac{1}{5} $.
We can not find the value of $ {{\tan }^{-1}}\dfrac{1}{5} $ by using any formula, we have to use the calculator.

Note:
We can transform $ {{\tan }^{-1}}\dfrac{2}{3} $ into many other forms, but we can not find the exact numerical value of $ {{\tan }^{-1}}\dfrac{2}{3} $ by using any formula, $ {{\tan }^{-1}}\dfrac{2}{3} $ is a irrational number. We can write $ {{\tan }^{-1}}\dfrac{2}{3} $ as $ \dfrac{\pi }{2}-{{\cot }^{-1}}\dfrac{2}{3} $ . We know the formula that $ {{\tan }^{-1}}x+{{\cot }^{-1}}x $ is always equal to $ \dfrac{\pi }{2} $
Always remember the range of $ {{\tan }^{-1}}x $ is from $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ . If we have to find the value of $ {{\tan }^{-1}}-1 $ then answer would be $ -\dfrac{\pi }{4} $ not $ \dfrac{3\pi }{4} $ .