Question

Solve the following for $x$:
 $3{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi $

Answer 1

Hint: we are asked to solve $3{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi $. Here, we will convert the equation in one trigonometric function using identity and then we will simplify to obtain the required answer.
Formula to be used:
The required trigonometric identity that is used to solve the given problem is as follows.
$\tan x = \cot \left( {\dfrac{\pi }{2} - x} \right)$

Complete step by step answer:
The given equation is
$3{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi $
$ \Rightarrow 2{\tan ^{ - 1}}\left( x \right) + {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi $
Before getting into next step, we shall consider ${\tan ^{ - 1}}\left( x \right) = p$
Let ${\tan ^{ - 1}}\left( x \right) = p$. Then, $\tan p = x$.
$\tan p = \cot \left( {\dfrac{\pi }{2} - p} \right)$ (Here we have substituted the trigonometric identity $\tan x = \cot \left( {\dfrac{\pi }{2} - x} \right)$)
$ \Rightarrow x = \cot \left( {\dfrac{\pi }{2} - p} \right)$ (We have substituted $\tan p = x$)
$ \Rightarrow {\cot ^{ - 1}}x = \dfrac{\pi }{2} - p$
$ \Rightarrow {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = p + \dfrac{\pi }{2} - p$ (Here we have added both the inverse of tangent and cotangent)
$ \Rightarrow {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \dfrac{\pi }{2}$ ……..$\left( 1 \right)$
Now, we shall get into our solution.
$3{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi $
$ \Rightarrow 2{\tan ^{ - 1}}\left( x \right) + {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi $
$ \Rightarrow 2{\tan ^{ - 1}}\left( x \right) + \dfrac{\pi }{2} = \pi $ (Here we have substituted the equation $\left( 1 \right)$)
$ \Rightarrow 2{\tan ^{ - 1}}\left( x \right) = \pi - \dfrac{\pi }{2}$
$ \Rightarrow 2{\tan ^{ - 1}}\left( x \right) = \dfrac{\pi }{2}$
$ \Rightarrow {\tan ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} \times \dfrac{1}{2}$
$ \Rightarrow {\tan ^{ - 1}}\left( x \right) = \dfrac{\pi }{4}$
$ \Rightarrow x = \tan \dfrac{\pi }{4}$
$ \Rightarrow x = 1$ (We know that $\tan \dfrac{\pi }{4} = 1$)
Hence $x = 1$ is the desired solution for the given equation.

Note:
    Here students should not that $ \Rightarrow {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \dfrac{\pi }{2}$ this expression is also an identity. So we can directly use it to simplify the calculation.