Question

$ta{n}^{-1}\left(\frac{c_{1}x-y}{c_{1}y+x}\right)+ta{n}^{-1}\left(\frac{c_2-c_1}{1+c_{2^{C}1}}\right)+..+ta{n}^{-1}\left(\frac{1}{c_n}\right)=$

• A. $ta{n}^{-1}\left(\frac{2x}{y}\right)$
• B. $ta{n}^{-1}\left(xy\right)$
• C. $ta{n}^{-1}\left(\frac{x}{y}\right)$
• D. $ta{n}^{-1}\left(y/x\right)$

Answer 1

The correct option is C $ta{n}^{-1}\left(\frac{x}{y}\right)$

$ta{n}^{-1}\left(\frac{c_{1}x-y}{c_{1}y+x}\right)$
$=ta{n}^{-1}\left(\frac{\frac{x}{y}-\frac{1}{c_1}}{1+\frac{x}{y{c}_1}}\right)$
$=ta{n}^{-1}\left(\frac{x}{y}\right)-ta{n}^{-1}\left(\frac{1}{c_1}\right)$

$ta{n}^{-1}\left(\frac{c_2-c_1}{1+c_2{c}_1}\right)=ta{n}^{-1}\left(\frac{1}{c_1}\right)-ta{n}^{-1}\left(\frac{1}{c_2}\right)$
Summing up we get.
$ta{n}^{-1}\left(\frac{x}{y}\right)-ta{n}^{-1}\left(\frac{1}{c_1}\right)+ta{n}^{-1}\left(\frac{1}{c_1}\right)-ta{n}^{-1}\left(\frac{1}{c_2}\right)...ta{n}^{-1}\left(\frac{1}{c_n}\right)$
Rest of the terms get eliminate and we are left with
$ta{n}^{-1}\left(\frac{x}{y}\right)$