Question

The sum of series ${\sec }^{-1}\sqrt{2}+{\sec }^{-1}\frac{\sqrt{10}}{3}+{\sec }^{-1}\frac{\sqrt{50}}{7}+...+{\sec }^{-1}\sqrt{\frac{(n^2+1)(n^2-2n+2)}{(n^2-n+1{)}^2}}$

• A. ${\tan }^{-1}1$
• B. ${\tan }^{-1}n$
• C. ${\tan }^{-1}(n+1)$
• D. ${\tan }^{-1}(n-1)$

Answer 1

The correct option is B ${\tan }^{-1}n$

$\text{Solve:}$

$\text{Given,}$

${\text{Sec}}^{-1}\sqrt{2}+{\sec }^{-1}\frac{\sqrt{10}}{3}+{\sec }^{-1}\frac{\sqrt{50}}{7}+\cdots +{\sec }^{-1}\sqrt{\frac{(n^2+1)(n^2-2n+2)}{{(n^2-n+1)}^2}}$

$\text{so, the}{n}_{th}\text{term,}{T}_n={\sec }^{-1}\sqrt{\frac{(n^2+1)(n^2-2n+2)}{{(n^2-n+1)}^2}}$

$\Rightarrow \sec T_n=\sqrt{\frac{(n^2+1)(n^2-2n+2)}{{(n^2-n+1)}^2}}$

$\Rightarrow {\sec }^2{T}_n=\frac{(n^2+1)(n^2-2n+2)}{{(n^2-n+1)}^2}$

$\Rightarrow {\sec }^2{T}_n=\frac{n^4-2n^3+2n^2+n^2-2n+2}{{(n^2-n+1)}^2}$

$\Rightarrow {\sec }^2{T}_n=\frac{(n^4+n^2+1-(2h)\left(n^2\right)+2n^2-2n+1}{{(n^2-n+1)}^2}$

$\Rightarrow {\tan }^2{T}_n+1=\frac{{(n^2+1-n)}^2+1}{{(n^2-n+1)}^2}=1+\frac{1}{{(n^2-n+1)}^2}$

$=>\tan T_n=\frac{1}{n^2-n+1}$

$\Rightarrow \tan T_n=\frac{n-(n-1)}{1+n(n-1)}$

$\Rightarrow \tan t_n={\tan }^{-1}n-{\mathrm{san}}^{-1}(n-1)$

$\Rightarrow \therefore \mathrm{sum},S=T_1+T_2+\cdots +T_n$

$\Rightarrow S={\tan }^{-1}n$

So, $ta{n}^{-1}n$ is the required sum of

series