Question

The number of solutions of ${\tan }^{-1}\left(\sqrt{x\left(x+1\right)}\right)+{\sin }^{-1}\left(\sqrt{x^2+x+1}\right)=\frac{\pi }{2}$

Answer 1

Find the root of the equation

Given, ${\tan }^{-1}\left(\sqrt{x\left(x+1\right)}\right)+{\sin }^{-1}\left(\sqrt{x^2+x+1}\right)=\frac{\pi }{2}$

$$\begin{gathered} \Rightarrow {\tan }^{-1}\left(\sqrt{x\left(x+1\right)}\right)=\frac{\pi }{2}-{\sin }^{-1}\left(\sqrt{x^2+x+1}\right) \\ \Rightarrow {\tan }^{-1}\left(\sqrt{x\left(x+1\right)}\right)={\cos }^{-1}\left(\sqrt{x^2+x+1}\right).........\left(i\right) \end{gathered}$$

Let,

$$\begin{gathered} \tan \left(\theta \right)=\sqrt{x\left(x+1\right)} \\ \Rightarrow \cos \left(\theta \right)=\frac{1}{\sqrt{x^2+x+1}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{x^2+x+1}}\right)...........\left(ii\right) \end{gathered}$$

Now using $\left(ii\right)$ in $\left(i\right)$ we get

$$\begin{gathered} {\cos }^{-1}\left(\frac{1}{\sqrt{x^2+x+1}}\right)={\cos }^{-1}\left(\sqrt{x^2+x+1}\right) \\ \Rightarrow \frac{1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1} \\ \Rightarrow x^2+x+1=1 \\ \Rightarrow x^2+x=0 \\ \Rightarrow x\left(x+1\right)=0 \\ x=0,-1 \end{gathered}$$

Hence, there are $2$ solutions for the given expression.