SELECT * FROM question_mgmt as q WHERE id=981 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 sec 1

Question: Find the principal value of \begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\right)\end{align}

Answer:

\begin{align} Let \;\; sec^{-1}\left(\frac{2}{\sqrt3}\right)=y \;\;Then\;\; sec y = \frac{2}{\sqrt3} = sec\left(\frac{\pi}{6}\right)\end{align}

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

 \begin{align} \left[0,\pi\right] - \left(\frac{\pi}{2}\right) and \;\;sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt3}\end{align}

Therefore, the principal value of

 \begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\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!=981 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!=981 ORDER BY last_viewed_on desc limit 0,10

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