SELECT * FROM question_mgmt as q WHERE id=982 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=6 AND chapterId=89 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th mathematics Inverse Trigonometric Functions find the principal value of begin align cot 1

Question: Find the principal value of \begin{align} cot^{-1}\left(\sqrt3\right)\end{align}

Answer:

\begin{align} Let \;\; cot^{-1}\left(\sqrt3\right)=y. \;\;Then,\;\; cot y = \sqrt3 = cot\left(\frac{\pi}{6}\right).\end{align}

We know that the range of the principal value branch of cot−1 is 

\begin{align} \left(0, \pi\right) and \;\;cot\left(\frac{\pi}{6}\right) = \sqrt3.\end{align}

 
Therefore, the principal value of
\begin{align} cot^{-1}\left(\sqrt 3\right) is \frac{\pi}{6}\end{align}


SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='6' AND ex_no!=0 AND status=1 and id!=982 ORDER BY views desc, last_viewed_on desc limit 0,10
SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='6' AND ex_no!=0 AND status=1 and id!=982 ORDER BY last_viewed_on desc limit 0,10

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  • 4 months ago

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