MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)\) is
- A \(\frac{4 \pi}{3}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{2 \pi}{4}\)
- D \(\frac{3 \pi}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \)
\( =\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3} \times \frac{1}{5}}\right)+\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7} \times \frac{1}{8}}\right) \)
\( =\tan ^{-1}\left(\frac{4}{7}\right)+\tan ^{-1}\left(\frac{3}{11}\right) \)
\( =\tan ^{-1} \frac{\frac{4}{7}+\frac{3}{11}}{\left(1-\frac{4}{7} \times \frac{3}{11}\right)}=\tan ^{-1}\left(\frac{65}{65}\right)=\tan ^{-1}(1)\) \(=\frac{\pi}{4}\)
\( =\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3} \times \frac{1}{5}}\right)+\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7} \times \frac{1}{8}}\right) \)
\( =\tan ^{-1}\left(\frac{4}{7}\right)+\tan ^{-1}\left(\frac{3}{11}\right) \)
\( =\tan ^{-1} \frac{\frac{4}{7}+\frac{3}{11}}{\left(1-\frac{4}{7} \times \frac{3}{11}\right)}=\tan ^{-1}\left(\frac{65}{65}\right)=\tan ^{-1}(1)\) \(=\frac{\pi}{4}\)
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