First slide

Introduction to ITF

Question

The greater of the two angles $A = 2 \tan^{- 1} (2 \sqrt{2} - 1)$ and $B = 3 \sin^{- 1} \frac{1}{3} + \sin^{- 1} \frac{3}{5}$ is

Moderate

A

B
B

A
C

C
D

none of these

Solution

$\begin{array}{l} A = 2 \tan^{- 1} (2 \sqrt{2} - 1) = 2 \tan^{- 1} (1 .828) \\ ∴ A > 2 \tan^{- 1} \sqrt{3} (\sqrt{3} = 1 .732 < 1 .828) \end{array}$
$\Rightarrow A > \frac{2 π}{3}$ ……. (1)
We have $\sin^{- 1} \frac{1}{3} < \sin^{- 1} \frac{1}{2} = \frac{π}{6}$
$\Rightarrow 3 \sin^{- 1} \frac{1}{3} < \frac{π}{2}$
Using $\sin 3 θ = 3 \sin θ - 4 \sin^{3} θ$ ,
we have, $\sin^{- 1} \frac{1}{3} = \sin^{- 1} (3 \times \frac{1}{3} - 4 {(\frac{1}{3})}^{3})$
$\begin{array}{l} = \sin^{- 1} (\frac{23}{27}) = \sin^{- 1} (0 .852) \\ ∴ 3 \sin^{- 1} \frac{1}{3} < \sin^{- 1} (\frac{\sqrt{3}}{2}) (∵ \frac{\sqrt{3}}{2} = 0 .868 > 0 .852) \end{array}$
i.e., $3 \sin^{- 1} \frac{1}{3} < \frac{π}{3}$ …….. (2)
Also, $\sin^{- 1} \frac{3}{5} = \sin^{- 1} (0 .6) < \sin^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{3}$
$\begin{array}{l} ∴ \sin^{- 1} \frac{3}{5} < \frac{π}{3} \\ ∴ B = 3 \sin^{- 1} \frac{1}{3} + \sin \frac{3}{5} < \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3} \frac{2 π}{3} \end{array}$
$∴ B < \frac{2 π}{3}$ ……….. (3)
From (1) and (3), A > B

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