SELECT * FROM question_mgmt as q WHERE id=993 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 value of if begin align tan 1 sqrt3

Question: Find the value of if \begin{align} tan^{-1}\sqrt3 - sec^{-1}(-2)\end{align}
is equal to
\begin{align} (A) \pi \;\;(B) -\frac{\pi}{3}\;\; (C) \frac{\pi}{3} \;\;(D) \frac{2\pi}{3}\end{align}

Answer:

\begin{align} Let \;\; tan^{-1}(\sqrt 3)=x. \;\;Then\;\; tan x = \sqrt 3 = tan\left(\frac{\pi}{3}\right).\end{align}
We know that the range of the principal value branch of tan−1 is
  \begin{align} \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\end{align}
  \begin{align}  \therefore tan^{-1}\sqrt3=\frac{\pi}{3}\end{align}
\begin{align} Let \;\; sec^{-1}\left(-2\right)=y \;\;Then\;\; sec y = -2 = -sec\left(\frac{\pi}{3}\right) = sec\left(\pi - \frac{\pi}{3}\right) = sec\left(\frac{2\pi}{3}\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)\end{align}
  \begin{align}  \therefore sec^{-1}({-2})=\frac{2\pi}{3}\end{align}
Hence, \begin{align} tan^{-1}\sqrt3 - sec^{-1}(-2)\end{align} 
\begin{align} =\frac{\pi}{3} - \frac{2\pi}{3}=-\frac{\pi}{3}\end{align} 
 


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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!=993 ORDER BY last_viewed_on desc limit 0,10

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