MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)\) is
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right) \\ & =\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2} \cdot \frac{1}{5}}\right) \\ & =\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{7}{9}\right) \\ & =\tan ^{-1}\left(\frac{\frac{1}{8}+\frac{7}{9}}{1-\frac{1}{8} \cdot \frac{7}{9}}\right) \\ & =\tan ^{-1}\left(\frac{\frac{72}{65}}{\frac{72}{9}}\right) \\ & =\tan ^{-1}(1) \\ & =\frac{\pi}{4}\end{aligned}\)
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