MHT CET · Maths · Inverse Trigonometric Functions
\(\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}\) has the value
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{6}\)
- C \(\frac{\pi}{2}\)
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8} \\ & =\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3} \times \frac{1}{5}}\right)+\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7} \times \frac{1}{8}}\right) \\ & =\tan ^{-1}\left(\frac{4}{7}\right)+\tan ^{-1}\left(\frac{3}{11}\right) \\ & =\tan ^{-1}\left(\frac{\frac{4}{7}+\frac{3}{11}}{1-\frac{4}{7} \times \frac{3}{11}}\right)\end{aligned}\)
\(=\tan ^{-1}\left(\frac{65}{65}\right)=\tan ^{-1}(1)=\frac{\pi}{4}\)
\(=\tan ^{-1}\left(\frac{65}{65}\right)=\tan ^{-1}(1)=\frac{\pi}{4}\)
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