ExamBro
ExamBro
TS EAMCET · Maths · Inverse Trigonometric Functions

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

  1. A \(\operatorname{Tan}^{-1} \frac{9}{10}\)
  2. B \(\operatorname{Tan}^{-1} \frac{18}{19}\)
  3. C \(\operatorname{Tan}^{-1} \frac{3}{191}\)
  4. D \(\operatorname{Tan}^{-1} \frac{6}{205}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\operatorname{Tan}^{-1} \frac{9}{10}\)

Step-by-step Solution

Detailed explanation

\operatorname{Tan}^{-1} \frac{3}{5}+\operatorname{Tan}^{-1} \frac{6}{41} = \operatorname{Tan}^{-1} \left( \frac{\frac{3}{5}+\frac{6}{41}}{1-\frac{3}{5} \cdot \frac{6}{41}} \right) = \operatorname{Tan}^{-1} \left( \frac{123+30}{205-18} \right) = \operatorname{Tan}^{-1}…