TS EAMCET · Maths · Inverse Trigonometric Functions
\(2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)\) is equal to
- A \(\tan ^{-1}\left(\frac{49}{29}\right)\)
- B \(\frac{\pi}{2}\)
- C 0
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & 2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\tan ^{-1}\left(\frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{9}\right)}\right) +\tan ^{-1}\left(\frac{1}{7}\right)\\ & =\tan ^{-1}\left(\frac{3}{4}\right)+\tan…
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