Question

Let $h\left(x\right)=\sec \left\{\frac{{\tan }^{-1}\left(\cos \right({\sin }^{-1}x\left)\right)+{\cot }^{-1}\left(\sin \right({\cos }^{-1}x\left)\right)}{3}\right\},$
where $x\in [-1,1],$ then which of the following option(s) is(are) correct?

• A. $\sum _{r=1}^{10}h\left(\frac{1}{r}\right)=\sum _{r=10}^{14}h\left(\frac{1}{r}\right)$
• B. $\sum _{r=1}^{10}h\left(\frac{1}{r}\right)=2\sum _{r=10}^{14}h\left(\frac{1}{r}\right)$
• C. $\sum _{r=1}^{10}h\left(\frac{1}{r}\right)=\sum _{r=11}^{20}h\left(\frac{1}{r}\right)$
• D. $2\sum _{r=1}^{10}h\left(\frac{1}{r}\right)=\sum _{r=11}^{20}h\left(\frac{1}{r}\right)$

Answer 1

Answer: (B), (C)

$\sum _{r=1}^{10}h\left(\frac{1}{r}\right)=\sum _{r=11}^{20}h\left(\frac{1}{r}\right)$

Given : $h\left(x\right)=\sec \left\{\frac{{\tan }^{-1}\left(\cos \right({\sin }^{-1}x\left)\right)+{\cot }^{-1}\left(\sin \right({\cos }^{-1}x\left)\right)}{3}\right\},-1\le x\le 1$
$=\sec \left\{\frac{{\tan }^{-1}\left(\cos \right({\cos }^{-1}\sqrt{1-x^2}\left)\right)+{\cot }^{-1}\left(\sin \right({\sin }^{-1}\sqrt{1-x^2}\left)\right)}{3}\right\}$
$h\left(x\right)=\sec \left\{\frac{{\tan }^{-1}\sqrt{1-x^2}+{\cot }^{-1}\sqrt{1-x^2}}{3}\right\}$
$=\sec \frac{\pi }{6}(\because {\tan }^{-1}y+{\cot }^{-1}y=\frac{\pi }{2})$
$\Rightarrow h\left(x\right)=\frac{2}{\sqrt{3}}$ (constant function)

$\Rightarrow \sum _{r=1}^{10}h\left(\frac{1}{r}\right)=\frac{2}{\sqrt{3}}\times 10=\frac{20}{\sqrt{3}}$
$\Rightarrow \sum _{r=10}^{14}h\left(\frac{1}{r}\right)=\frac{2}{\sqrt{3}}\times 5=\frac{10}{\sqrt{3}}$
$\Rightarrow \sum _{r=11}^{20}h\left(\frac{1}{r}\right)=\frac{2}{\sqrt{3}}\times 10=\frac{20}{\sqrt{3}}$