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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

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

1 Answer

Step 1:
$A=2\tan^{-1}(2\sqrt 2-1)$
$\sqrt 2=1.414$
$\;\;=2\tan^{-1}(2\times 1.414-1)$
$\;\;=2\tan^{-1}(1.828)$
$\tan 60^{\large\circ}=\sqrt 3$
$\Rightarrow 1.732$
$\tan^{-1}(1.732)=60^{\large\circ}$
$\;\;=2\times ( >60^{\large\circ})$
$\;\;=>120^{\large\circ}$
Step 2:
$B=3\sin^{-1}\big(\large\frac{1}{3}\big)$$+\sin^{-1}\big(\large\frac{3}{5}\big)$
$\;\;=\sin^{-1}\big[3\times \large\frac{1}{3}$$-4\big(\large\frac{1}{3}\big)^3\big]+$$\sin^{-1}\big(\large\frac{3}{5}\big)$
$\;\;=\sin^{-1}\big(\large\frac{23}{27}\big)$$+\sin^{-1}\big(\large\frac{3}{5}\big)$
$\;\;=\sin^{-1}(0.852)+\sin^{-1}(0.60)$
$\;\;= (< 60^{\large\circ})+( < 45^{\large\circ})$
$\;\;=< 105^{\large\circ}$
$\Rightarrow A > B$
Hence (a) is the correct answer.
answered Oct 21, 2013 by sreemathi.v
edited Mar 24, 2014 by thanvigandhi_1
 

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