Question

if f(x)=${\tan }^{-1}$(x)than f(x)+f(y) is equal to:

• A. ${\tan }^{-1}x-{\tan }^{-1}y$=${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$
  
• B. ${\tan }^{-1}x+{\tan }^{-1}y$=${\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$
• C. ${\tan }^{-1}x+{\tan }^{-1}y$=${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$
• D. None of these

Answer 1

The correct option is C ${\tan }^{-1}x+{\tan }^{-1}y$=${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

Given that ,$f\left(x\right)={\tan }^{-1}x$ ….(1)

Put x=y in equation (1) we get

$f\left(y\right)={\tan }^{-1}y$ ….(2)

Add equation (1) and (2)

$f\left(x\right)+f\left(y\right)={\tan }^{-1}x+{\tan }^{-1}y$

Using inverse trigonometry properties, we get

${\tan }^{-1}x+{\tan }^{-1}y$ $={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

This is the answer