TS EAMCET · Maths · Inverse Trigonometric Functions
\(\operatorname{Tan}^{-1} \frac{3}{5}+\operatorname{Tan}^{-1} \frac{6}{41}+\operatorname{Tan}^{-1} \frac{9}{191}=\)
- A \(\operatorname{Tan}^{-1} \frac{9}{10}\)
- B \(\operatorname{Tan}^{-1} \frac{18}{19}\)
- C \(\operatorname{Tan}^{-1} \frac{3}{191}\)
- D \(\operatorname{Tan}^{-1} \frac{6}{205}\)
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}…
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