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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The greater of the angles $A=2\tan^{-1}(2\sqrt 2-1),B=3\sin^{-1}\big(\large\frac{1}{3}\big)$$+\sin^{-1}(\large\frac{3}{5})$ is

$(A)\;A < B\qquad(B)\;A=B\qquad(C)\;A > B\qquad(D)\;None\;of\;these$

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1 Answer

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Step 1:
$A=2\tan^{-1}(2\sqrt 2-1)$
$\sqrt 2=1.414$
$\;\;=2\tan^{-1}(2\times 1.414-1)$
$\tan 60^{\large\circ}=\sqrt 3$
$\Rightarrow 1.732$
$\;\;=2\times ( >60^{\large\circ})$
Step 2:
$\;\;=\sin^{-1}\big[3\times \large\frac{1}{3}$$-4\big(\large\frac{1}{3}\big)^3\big]+$$\sin^{-1}\big(\large\frac{3}{5}\big)$
$\;\;= (< 60^{\large\circ})+( < 45^{\large\circ})$
$\;\;=< 105^{\large\circ}$
$\Rightarrow A > B$
Hence (a) is the correct answer.
answered Oct 21, 2013 by sreemathi.v
edited Mar 24, 2014 by thanvigandhi_1

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