MHT CET · Maths · Inverse Trigonometric Functions
\(2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=\)
- A \(0\)
- B \(2\)
- C \(1\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\(2 \tan ^{-1} \frac{1}{3}=\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{3}}{1-\frac{1}{3} \times \frac{1}{3}}\right)\)
\(=\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right)=\tan ^{-1} \frac{3}{4}\)
Now \(2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{3}{4}\right)\) \(=0\)
\(=\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right)=\tan ^{-1} \frac{3}{4}\)
Now \(2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{3}{4}\right)\) \(=0\)
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