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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions2 Marks
Question
Evaluate the following:
$\cot^{-1}\Big(\cot\frac{19\pi}{6}\Big)$
✓

Answer

$\cot^{-1}\Big(\cot\frac{19\pi}{6}\Big)$
$=\cot^{-1}\Big[\cot\Big(3\pi+\frac{\pi}{6}\Big)\Big]$
$=\cot^{-1}\Big(\cot\frac{\pi}{6}\Big)$
$=\frac{\pi}{6}$

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