$\operatorname{Tan}^{-1} \frac{3}{5}+\operatorname{Tan}^{-1} \frac{6}{41}+\operatorname{Tan}^{-1} \frac{9}{191}=$

• Correct answer
  A

  $\operatorname{Tan}^{-1} \frac{9}{10}$
• B

  $\operatorname{Tan}^{-1} \frac{18}{19}$
• C

  $\operatorname{Tan}^{-1} \frac{3}{191}$
• D

  $\operatorname{Tan}^{-1} \frac{6}{205}$

Hence, the correct answer is option: 'A'.