Question

The value of $ta{n}^{-1}\frac{1}{2}+ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{7}{8}$ is

• A. tan^-1 7/8
• B. ^cot-1 15
• C. tan^-1 15
• D. tan^-1 25/24

Answer 1

The correct option is C

tan^-1 15

$ta{n}^{-1}\frac{1}{2}+ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{7}{8}$
$=ta{n}^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}\right)+ta{n}^{-1}\frac{7}{8}$
$=ta{n}^{-1}\left(\frac{5/6}{5/6}\right)+=ta{n}^{-1}\frac{7}{8}=ta{n}^{-1}\left(1\right)+ta{n}^{-1}\frac{7}{8}$
$=ta{n}^{-1}\left(\frac{1+\frac{7}{8}}{1-\frac{7}{8}}\right)=ta{n}^{-1}\left(15\right)$