Question

If $tan(x+y)=33andx=ta{n}^{-1}3$ , then $y$ will be

• A. $0.3$
• B. $ta{n}^{-1}\left(1.3\right)$
• C. $ta{n}^{-1}\left(0.3\right)$
• D. $ta{n}^{-1}\left(\frac{1}{18}\right)$

Answer 1

The correct option is C $ta{n}^{-1}\left(0.3\right)$

$y={\tan }^{-1}\left(33\right)-{\tan }^{-1}\left(3\right)$

$={\tan }^{-1}\left(\frac{33-3}{1+99}\right)$

$={\tan }^{-1}\left(\frac{30}{100}\right)$

$={\tan }^{-1}\left(\frac{3}{10}\right)$