Question

If $a_{1}, \, a_{2}, \, a_{3}$ are in arithmetic progression and $d$ is the common difference, then $\left(tan\right)^{- 1} \, \left(\frac{d}{1 + a_{1} a_{2}}\right) + \left(tan\right)^{- 1} ⁡ \, \left(\frac{d}{1 + a_{2} a_{3}}\right) =$

• A. $\left(tan\right)^{- 1} \, \left(\frac{2 d}{1 + a_{1} a_{3}}\right)$
• B. $\left(tan\right)^{- 1} \, \left(\frac{d}{1 + a_{1} a_{3}}\right)$
• C. $\left(tan\right)^{- 1} \, \left(\frac{2 d}{1 + a_{2} a_{3}}\right)$
• D. $\left(tan\right)^{- 1} \, \left(\frac{2 d}{1 - a_{1} a_{3}}\right)$

Answer 1

Answer: (A)

Now, $\tan ^{-1}\left(\frac{d}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{d}{1+a_{2} a_{3}}\right)$
$=\tan ^{-1}\left(\frac{a_{2}-a_{1}}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{a_{3}-a_{2}}{1+a_{2} a_{3}}\right)$
$\left(\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}\right)$
$=\tan ^{-1} a_{2}-\tan ^{-1} a_{1}+\tan ^{-1} a_{3}-\tan ^{-1} a_{2}$
$=\tan ^{-1} a_{3}-\tan ^{-1} a$
$=\tan ^{-1}\left(\frac{a_{3}-a_{1}}{1+a_{1} a_{3}}\right)$
$=\tan ^{-1}\left(\frac{\left(a_{3}-a_{2}\right)+\left(a_{2}-a_{1}\right)}{1+a_{1} a_{3}}\right)$
$=\tan ^{-1}\left(\frac{2 d}{1+a_{1} a_{3}}\right)$