MHT CET · Maths · Inverse Trigonometric Functions
\( \sin ^{-1}\left[\sin \left(-600^{\circ}\right)\right]+\cot ^{-1}(-\sqrt{3})= \)
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{7 \pi}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{6}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \sin ^{-1}\left[\sin (-600)^{\circ}\right]+\cot ^{-1}(\sqrt{3}) \\
& =\sin ^{-1}\left[-\sin \left(360^{\circ}+180^{\circ}+60^{\circ}\right)\right]+\left[-\cot ^{-1}(\sqrt{3})\right] \\
& =\sin ^{-1}\left[\sin \left(60^{\circ}\right)\right]+\left[-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right] \\
& =\sin ^{-1}\left[\sin 60^{\circ}\right]-\frac{\pi}{6}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}
\end{aligned}
\)
\begin{aligned}
& \sin ^{-1}\left[\sin (-600)^{\circ}\right]+\cot ^{-1}(\sqrt{3}) \\
& =\sin ^{-1}\left[-\sin \left(360^{\circ}+180^{\circ}+60^{\circ}\right)\right]+\left[-\cot ^{-1}(\sqrt{3})\right] \\
& =\sin ^{-1}\left[\sin \left(60^{\circ}\right)\right]+\left[-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right] \\
& =\sin ^{-1}\left[\sin 60^{\circ}\right]-\frac{\pi}{6}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}
\end{aligned}
\)
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