Question

Let $h(x)=\tan \left(\frac{{\cos }^{-1}\left(\sin \left({\cos }^{-1}x\right)\right)+{\sin }^{-1}\left(\cos \left({\sin }^{-1}x\right)\right)}{2}\right)$ then find the value of $\sum _{x=1}^{7}h\left(\frac{x}{8}\right)$.

Answer 1

Answer: 7

${\cos }^{-1}\left(\sin \left({\cos }^{-1}x\right)\right)+{\sin }^{-1}\left(\cos \left({\sin }^{-1}x\right)\right)={\cos }^{-1}\sqrt{1-x^2}+{\sin }^{-1}\sqrt{1-x^2}=\frac{\pi }{2}$
$\therefore h(x)=1$
$\therefore \sum _{x=1}^{7}h\left(\frac{x}{8}\right)=7$