JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If the inverse trigonometric functions take principal values, then \(\cos ^{-1}\left(\frac{3}{10} \cos \left(\tan ^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5} \sin \left(\tan ^{-1}\left(\frac{4}{3}\right)\right)\right)\) is equal to
- A \(0\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
Let \(\tan ^{-1} \frac{4}{3}=\theta \Rightarrow \tan \theta=\frac{4}{3}\) \(E =\cos ^{-1}\left(\frac{3}{10} \cos \theta+\frac{2}{5} \sin \theta\right)\) \(=\cos ^{-1}\left(\frac{3}{10} \times \frac{3}{5}+\frac{2}{5} \cdot \frac{4}{5}\right)\)…
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