Question

If the expression $\arctan \left(\frac{1}{x}-\frac{x}{8}\right)+\arctan (a x)+\arctan (b x)=\frac{\pi}{2}(a, b \in R)$ is true $\forall x \in R _0$, then find the value of $4\left( a ^2+ b ^2\right)$.

Answer 1

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, $1=a x\left(\frac{1}{x}-\frac{x}{8}\right)+a b x^2+b x\left(\frac{1}{x}-\frac{x}{8}\right) \Rightarrow 1=a-\frac{a}{8} x^2+a b x^2+b-\frac{b}{8} x^2 \forall x \in R$
$1=(a+b)+x^2\left(a b-\left(\frac{a+b}{8}\right)\right)$
$\therefore $ On comparing, we get
$a+b=1$.......(1)
and $a b=\frac{a+b}{8}=\frac{1}{8}$......(2) (Using (1))
Now, $a^2+b^2+2 a b=1 \Rightarrow a^2+b^2+\frac{1}{4}=1$
Hence, $4\left(a^2+b^2\right)=3$.