Q51 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics
If $\tan ^{-1}\left[\frac{1}{1+1.2}\right]+\tan ^{-1}\left[\frac{1}{1+2.3}\right]+$ $\ldots+\tan ^{-1}\left[\frac{1}{1+n(n+1)}\right]=\tan ^{-1}[x]$, then $x$ is
equal to
- A$\frac{1}{n+1}$
- B$\frac{n}{n+1}$
- C$\frac{1}{n+2}$
- Correct answerD$\frac{n}{n+2}$
Solution
$\begin{aligned} & \text { Given } \tan ^{-1}\left(\frac{1}{1+1.2}\right)+\tan ^{-1}\left(\frac{1}{1+2.3}\right)+\ldots \\ & +\tan ^{-1}\left(\frac{1}{1+n(n+1)}\right)=\tan ^{-1}(x)\end{aligned}$
$\begin{aligned} \Rightarrow \tan ^{-1}\left(\frac{2-1}{1+1.2}\right) & +\tan ^{-1}\left(\frac{3-2}{1+2.3}\right)+\ldots \\ & +\tan ^{-1}\left(\frac{n+1-n}{1+n(n+1)}\right)=\tan ^{-1}(x)\end{aligned}$
$\begin{aligned} & \Rightarrow \tan ^{-1}(2)-\tan ^{-1}(1)+\tan ^{-1}(3)-\tan ^{-1}(2)+\ldots \\ & +\tan ^{-1}(\mathrm{n}+1)-\tan ^{-1}(\mathrm{n})=\tan ^{-1}(x)\end{aligned}$
$\Rightarrow \tan ^{-1}(\mathrm{n}+1)-\tan ^{-1}(1)=\tan ^{-1}(x)$
$\left\{\because \tan ^{-1}(A)-\tan ^{-1}(B)=\tan ^{-1}\left(\frac{A-B}{1+A \cdot B}\right)\right\}$
$\Rightarrow \tan ^{-1}\left(\frac{n}{n+2}\right)=\tan ^{-1}(x) \Rightarrow x=\frac{n}{n+2}$
Hence, the correct answer is option: 'D'.