Question

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

• A. ${\tan }^{-1}\frac{7}{8}$
• B. ${\cot }^{-1}15$
• C. ${\tan }^{-1}15$
• D. ${\tan }^{-1}\frac{25}{24}$

Answer 1

Answer: (C)

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