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Solve the following \(\tan^{-1} \sqrt{3} -sec^{-1} (-2)\):

\[ (A) \quad \pi \qquad (B) \quad -\frac{\pi}{3} \qquad (C) \quad \frac{\pi}{3} \qquad (D) \quad \frac{2\pi}{3} \qquad \]
Can you answer this question?
 
 

2 Answers

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Ans : B
\( \frac{\pi}{3} - \bigg( \pi-\frac{\pi}{3} \bigg) = \frac{-\pi}{3} \)
answered Feb 22, 2013 by thanvigandhi_1
 
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Toolbox:
  • The range of the principal value of $\tan^{-1}x$ is $\left [ -\large\frac{\pi}{2},\large \frac{\pi}{2} \right ]$
  • The range of the principal value of $\sec^{-1}x$ is $\left [0, \pi \right ] - (\large\frac{\pi}{2})$
  • $sec (\pi - \theta) = - sec \theta$
Let $\tan^{-1} \sqrt3 = x$
$\Rightarrow \tan x = \sqrt 3= tan\large\frac{\pi}{3}$
 
\( \therefore\), $x = \large\frac{\pi}{3} \; \epsilon \left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
Let $\sec^{-1} (-2) = y$
$ \Rightarrow \sec y = -2$
 
\( \therefore\), $y = -sec \large\frac{\pi}{2}$
$\Rightarrow sec y = sec (\pi -\large \frac{\pi}{2}) = sec(\large\frac{2\pi}{3})$
 
\(\therefore\), $y = \large\frac{2\pi}{3} \;\; \epsilon \;\left [0, \pi \right ] - (\large\frac{\pi}{2})$
$\tan^{-1} \sqrt3 - \sec^{-1} (-2) = x- y =\large \frac{\pi}{3} - \large\frac{2\pi}{3} = -\large\frac{\pi}{3}$
 
The correct answer is $(B): -\large\frac{\pi}{3}$

 

answered Mar 1, 2013 by balaji.thirumalai
edited Mar 15, 2013 by thanvigandhi_1
 
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