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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $a=\sin^{-1}\big(\large\frac{-1}{\sqrt 2}\big)+$$\cos^{-1}\big(\large\frac{-1}{2}\big),$$b=\tan^{-1}\big(-\sqrt 3\big)-\cot^{-1}\big(-\large\frac{1}{\sqrt 3}\big)$ then

$\begin{array}{1 1}(a)\;a-b=\large\frac{17\pi}{12}&(b)\;a+b=\large\frac{17\pi}{12}\\(c)\;a+b=\large\frac{7\pi}{12}&(d)\;a-b=\large\frac{\pi}{12}\end{array}$

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1 Answer

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Since $a=\sin^{-1}[\sin(-\large\frac{\pi}{4})]$$+\cos^{-1}[\cos\large\frac{2\pi}{3}]$
$\qquad\quad=\large\frac{-\pi}{4}+\frac{2\pi}{3}$
$\qquad\quad=\large\frac{5\pi}{12}$
$b=\tan^{-1}\big(\tan\big(-\large\frac{\pi}{3}\big)\big)$$-\cot^{-1}\big(\cot\big(\large\frac{2\pi}{3}\big)\big)$
$\;\;\;=-\large\frac{\pi}{3}-\frac{2\pi}{3}$
$\;\;\;=-\pi$
$a+b=\large\frac{5\pi}{12}$$-\pi$
$\qquad\;=-\large\frac{7\pi}{12}$
$a-b=\large\frac{5\pi}{12}$$+\pi$
$\qquad\;=\large\frac{17\pi}{12}$
Hence (a) is the correct option.
answered Oct 11, 2013 by sreemathi.v
edited Mar 22, 2014 by sharmaaparna1
 

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