Question

If the expression $\arctan \left(\frac{1}{x}-\frac{x}{8}\right)+\arctan (ax)+\arctan (bx)=\frac{\pi }{2}(a,b\in R)$ is true $\forall x\in R_0$, then find the value of $4\left(a^2+b^2\right)$.

Answer 1

Answer: 3

Using ${\tan }^{-1}\alpha +{\tan }^{-1}\beta +{\tan }^{-1}\gamma ={\tan }^{-1}\left(\frac{\alpha +\beta +\gamma -\alpha \beta \gamma }{1-\sum \alpha \beta }\right)$ In L.H.S. we get
${\tan }^{-1}\left(\frac{c+ax+bx-c\cdot ax\cdot bx}{1-c(ax)-(ax)(bx)-(bx)(c)}\right)=\frac{\pi }{2}$
where $c=\left(\frac{1}{x}-\frac{x}{8}\right)$
Now, $1=ax\left(\frac{1}{x}-\frac{x}{8}\right)+ab{x}^2+bx\left(\frac{1}{x}-\frac{x}{8}\right)\Rightarrow 1=a-\frac{a}{8}{x}^2+ab{x}^2+b-\frac{b}{8}{x}^2\forall x\in R$
$1=(a+b)+x^2\left(ab-\left(\frac{a+b}{8}\right)\right)$
$\therefore$ On comparing, we get
$a+b=1$.......(1)
and $ab=\frac{a+b}{8}=\frac{1}{8}$......(2) (Using (1))
Now, $a^2+b^2+2ab=1\Rightarrow a^2+b^2+\frac{1}{4}=1$
Hence, $4\left(a^2+b^2\right)=3$.