MHT CET · Maths · Inverse Trigonometric Functions
The value of \(2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{3}{8}\) is
- A \(\tan ^{-1}\left(\frac{42}{24}\right)\)
- B \(2 \tan ^{-1}\left(\frac{42}{24}\right)\)
- C \(\tan ^{-1}\left(\frac{24}{41}\right)\)
- D \(\tan ^{-1}\left(\frac{41}{12}\right)\)
Answer & Solution
Correct Answer
(D) \(\tan ^{-1}\left(\frac{41}{12}\right)\)
Step-by-step Solution
Detailed explanation
\(2 \tan ^{-1} \frac{1}{2} = \tan ^{-1} \left( \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} \right) = \tan ^{-1} \left( \frac{1}{1 - \frac{1}{4}} \right) = \tan ^{-1} \left( \frac{1}{\frac{3}{4}} \right) = \tan ^{-1} \frac{4}{3}\) \(\tan ^{-1} \frac{4}{3} + \tan ^{-1} \frac{3}{8} = \tan ^{-1} \left( \frac{\frac{4}{3} + \frac{3}{8}}{1 - \frac{4}{3} \cdot \frac{3}{8}} \right)\)
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