KCET · Maths · Inverse Trigonometric Functions
\(\tan ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}\right] +\sin ^{-1}\left[\cos \left(\sin ^{-} \frac{\sqrt{3}}{2}\right)\right]\) is equal to
- A 0
- B \(\frac{\pi}{6}\)
- C \(\frac{\pi}{3}\)
- D \(\pi\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\tan ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}\right] = \tan ^{-1}\left[\frac{1}{\sqrt{3}} \cdot 1\right] = \frac{\pi}{6}\) \(\sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right] = \sin ^{-1}\left[\cos \left(\frac{\pi}{3}\right)\right] = \sin ^{-1}\left[\frac{1}{2}\right] = \frac{\pi}{6}\)
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