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