MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)\) is
- A \(\frac{6}{17}\)
- B \(\frac{7}{16}\)
- C \(\frac{16}{7}\)
- D \(\frac{17}{6}\)
Answer & Solution
Correct Answer
(D) \(\frac{17}{6}\)
Step-by-step Solution
Detailed explanation
\(\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\tan \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right)\)
\(=\tan \tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}\right)=\tan \tan ^{-1}\left(\frac{17}{6}\right)=\frac{17}{6}\)
\(=\tan \tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}\right)=\tan \tan ^{-1}\left(\frac{17}{6}\right)=\frac{17}{6}\)
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