Question

If ${\tan }^{-1}\left(\frac{1}{1+2}\right)+{\tan }^{-1}\left(\frac{1}{1+\left(2\right)\left(3\right)}\right)+{\tan }^{-1}\left(\frac{1}{1+\left(3\right)\left(4\right)}\right)+...+{\tan }^{-1}\left(\frac{1}{1+\left(n\right)\left(n+1\right)}\right)={\tan }^{-1}\left(\theta \right)$. Then $\theta$ is equal

• A. $\frac{n}{n+1}$
• B. $\frac{n+1}{n+2}$
• C. $\frac{n+2}{n+1}$
• D. $\frac{n}{n+2}$

Answer 1

The correct option is D

$\frac{n}{n+2}$

Explanation for the correct option:

Resolve each term of the given sequence

$\begin{array}{c}\text{Here,}ta{n}^{-1}\left(\frac{1}{1+1·2}\right)=ta{n}^{-1}\left(\frac{2-1}{1+2·1}\right)\\ =ta{n}^{-1}2-ta{n}^{-1}1\end{array}\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x+y}}{{\displaystyle 1-xy}}\right)\mathrm{or}{\tan }^{-1}\mathrm{x}-{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left(\frac{{\displaystyle \mathrm{x}-\mathrm{y}}}{{\displaystyle 1+\mathrm{xy}}}\right)\right]$

$\begin{array}{l}\text{Similarly,}{\tan }^{-1}\left(\frac{1}{1+2·3}\right)={\tan }^{-1}3-{\tan }^{-1}2\\ \text{and}{\tan }^{-1}\left(\frac{1}{1+n(n+1)}\right)={\tan }^{-1}(n+1)-{\tan }^{-1}n\end{array}$

Therefore, ${\tan }^{-1}2-{\tan }^{-1}1+{\tan }^{-1}3-{\tan }^{-1}2+\mathrm{....}+{\tan }^{-1}(n-1)-{\tan }^{-1}n={\tan }^{-1}\theta$

$\begin{array}{c}{\text{or,\hspace{0.17em}\hspace{0.17em}tan}}^{-1}(n+1)-{\text{\hspace{0.17em}tan}}^{-1}1={\text{\hspace{0.17em}tan}}^{-1}\theta \\ {\text{or,\hspace{0.17em}\hspace{0.17em}tan}}^{-1}\left(\frac{n+1-1}{1+(n+1)\left(1\right)}\right)={\text{\hspace{0.17em}tan}}^{-1}\theta \\ \text{or,\hspace{0.17em}}\frac{n}{1+n+1}=\theta \\ \text{or,\hspace{0.17em}}\theta =\frac{n}{2+n}\end{array}$

Hence option (D) is the answer