Class 12 Mathematics - Chapter Inverse Trigonometric Functions NCERT Solutions | Find the value of if \begin{align} tan^{
Welcome to the NCERT Solutions for Class 12th Mathematics - Chapter Inverse Trigonometric Functions. This page offers a step-by-step solution to the specific question from Excercise 1 , Question 14: find the value of if begin align tan 1 sqrt3....
Question 14
Find the value of if \begin{align} tan^{-1}\sqrt3 - sec^{-1}(-2)\end{align}
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}
More Questions From Class 12 Mathematics - Chapter Inverse Trigonometric Functions
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