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. ${\left(tan\right)}^{-1}\left(\frac{2d}{1+a_1{a}_3}\right)$
• B. ${\left(tan\right)}^{-1}\left(\frac{d}{1+a_1{a}_3}\right)$
• C. ${\left(tan\right)}^{-1}\left(\frac{2d}{1+a_2{a}_3}\right)$
• D. ${\left(tan\right)}^{-1}\left(\frac{2d}{1-a_1{a}_3}\right)$

Answer 1

Answer: (A)

Now, ${\tan }^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{d}{1+a_2{a}_3}\right)$
$={\tan }^{-1}\left(\frac{a_2-a_1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{a_3-a_2}{1+a_2{a}_3}\right)$
$\left(\because {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\frac{x-y}{1+xy}\right)$
$={\tan }^{-1}{a}_2-{\tan }^{-1}{a}_1+{\tan }^{-1}{a}_3-{\tan }^{-1}{a}_2$
$={\tan }^{-1}{a}_3-{\tan }^{-1}a$
$={\tan }^{-1}\left(\frac{a_3-a_1}{1+a_1{a}_3}\right)$
$={\tan }^{-1}\left(\frac{\left(a_3-a_2\right)+\left(a_2-a_1\right)}{1+a_1{a}_3}\right)$
$={\tan }^{-1}\left(\frac{2d}{1+a_1{a}_3}\right)$