Question

Let $h\left(x\right)={\tan }^2\left\{\frac{2{\sin }^{-1}\left(\cos \right({\sin }^{-1}3x\left)\right)+2{\cos }^{-1}\left(\sin \right({\cos }^{-1}3x\left)\right)}{3}\right\};$ where $x\in [-\frac{1}{3},\frac{1}{3}],$ then the value of $\sum _{r=3}^{10}h\left(\frac{1}{r^2}\right)=$

Answer 1

Given : $h\left(x\right)={\tan }^2\left\{\frac{2{\sin }^{-1}\left(\cos \right({\sin }^{-1}3x\left)\right)+2{\cos }^{-1}\left(\sin \right({\cos }^{-1}3x\left)\right)}{3}\right\}$
$={\tan }^2\left\{\frac{2{\sin }^{-1}\left(\cos \right({\cos }^{-1}\sqrt{1-9x^2}\left)\right)+2{\cos }^{-1}\left(\sin \right({\sin }^{-1}\sqrt{1-9x^2}\left)\right)}{3}\right\}$
$={\tan }^2\left\{\frac{2{\sin }^{-1}\sqrt{1-9x^2}+2{\cos }^{-1}\sqrt{1-9x^2}}{3}\right\})$
$={\tan }^2\{\frac{2}{3}\times \frac{\pi }{2}\}(\because {\sin }^{-1}y+{\cos }^{-1}y=\frac{\pi }{2},|y|\le 1)$
$=3$ (constant function)
Also, $-\frac{1}{3}\le x\le \frac{1}{3}$
$\therefore \sum _{r=3}^{10}h\left(\frac{1}{r^2}\right)=8\times 3=24$