Q61 / 100 · BITSAT 2023 (Memory Based Paper 1) · Mathematics
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)$ is equal to
- A$\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$
- B$\frac{1}{2} \sin ^{-1}\left(\frac{3}{5}\right)$
- C$\frac{1}{2} \tan ^{-1}\left(\frac{3}{5}\right)$
- Correct answerD$\tan ^{-1}\left(\frac{1}{2}\right)$
Solution
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\tan ^{-1}\left\{\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right\}$
$=\tan ^{-1}\left\{\frac{9+8}{36-2}\right\}=\tan ^{-1}\left(\frac{1}{2}\right)$
$=\tan ^{-1}\left\{\frac{9+8}{36-2}\right\}=\tan ^{-1}\left(\frac{1}{2}\right)$
Hence, the correct answer is option: 'D'.