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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

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Correct option is B

A=2tan−1⁡(22−1)=2tan−1⁡(1.828)∴A>2tan−1⁡3(3=1.732<1.828)⇒A>2π3 ……. (1)We have sin−1⁡130.852i.e., 3sin−1⁡13<π3 …….. (2)Also, sin−1⁡35=sin−1⁡(0.6) B

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