Question

If $a_1,a_2,.....a_n$ is an AP with a common difference $d$, then $\tan \left[{\tan }^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{d}{1+a_2{a}_3}\right)+........+{\tan }^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)\right]$

• A. $\frac{(n-1)d}{a_1+a_n}$
• B. $\frac{(n-1)d}{1+a_1{a}_n}$
• C. $\frac{nd}{1+a_1{a}_n}$
• D. $\frac{(a_n-a_1)}{(a_n+a_1)}$

Answer 1

The correct option is B

$\frac{(n-1)d}{1+a_1{a}_n}$

Explanation for the correct option::

$$\begin{gathered} \tan \left[{\tan }^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{d}{1+a_2{a}_3}\right)+........+{\tan }^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)\right] \\ =\tan \left[{\tan }^{-1}\left(\frac{a_2-a_1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{a_3-a_2}{1+a_2{a}_3}\right)+........+{\tan }^{-1}\left(\frac{a_n-a_{n-1}}{1+a_{n-1}{a}_n}\right)\right] \\ =\tan \left[{\tan }^{-1}\left(a_2\right)-{\tan }^{-1}\left(a_1\right)+{\tan }^{-1}\left(a_3\right)-{\tan }^{-1}\left(a_2\right)+........+{\tan }^{-1}\left(a_n\right)-{\tan }^{-1}\left(a_{n-1}\right)\right] \\ =\tan \left[{\tan }^{-1}\left(a_n\right)-{\tan }^{-1}\left(a_1\right)\right] \\ =\tan \left[{\tan }^{-1}\left(\frac{a_n-a_1}{1+a_1{a}_n}\right)\right]\left[\because ta{n}^{-1}\left(A\right)-ta{n}^{-1}\left(B\right)=ta{n}^{-1}\left(\frac{A-B}{1+AB}\right)\right] \\ =\left(\frac{a_n-a_1}{1+a_1{a}_n}\right) \\ =\left(\frac{a_1+(n-1)d-a_{1=}}{1+a_1{a}_n}\right)\left[\because a_n=a_1+(n-1)d\right] \\ =\left(\frac{(n-1)d}{1+a_1{a}_n}\right) \end{gathered}$$

Hence, the correct option is (B).