Solveeit Logo

Question

Question: If A = 2 tan<sup>–1</sup>(2\(\sqrt{2}\)– 1) and B = 3 sin<sup>–1</sup>\(\left( \frac{1}{3} \right)\)...

If A = 2 tan–1(22\sqrt{2}– 1) and B = 3 sin–1(13)\left( \frac{1}{3} \right)+ sin–1(35)\left( \frac{3}{5} \right), then-

A

A = B

B

A < B

C

A > B

D

None of these

Answer

A > B

Explanation

Solution

We have A = 2 tan–1(22\sqrt{2}– 1) = 2 tan–1 (1.828) ̃ A > 2 tan–1 3\sqrt{3} ̃ A > 2π3\frac{2\pi}{3}

Next sin– 1(13)\left( \frac{1}{3} \right)< sin–1(12)\left( \frac{1}{2} \right) ̃ sin– 1(13)\left( \frac{1}{3} \right) < π6\frac{\pi}{6}

̃ sin– 113\frac{1}{3}< π2\frac{\pi}{2}

Also 3 sin– 1 (13)\left( \frac{1}{3} \right)= sin–1 [3.134(13)3]\left\lbrack 3.\frac{1}{3} - 4\left( \frac{1}{3} \right)^{3} \right\rbrack

= sin–1 (2327)\left( \frac{23}{27} \right)= sin–1 (0. 852)

̃ 3 sin–1 (13)\left( \frac{1}{3} \right)< sin–1 (32)\left( \frac{\sqrt{3}}{2} \right)̃ 3 sin–1(13)\left( \frac{1}{3} \right) <π3\frac{\pi}{3}

Further sin–1 (35)\left( \frac{3}{5} \right)= sin–1 (0. 6) < sin–1 (32)\left( \frac{\sqrt{3}}{2} \right)

̃ sin–1(35)\left( \frac{3}{5} \right)<π3\frac{\pi}{3}

Hence, B = 3 sin–1(13)\left( \frac{1}{3} \right)+ sin–1 (35)\left( \frac{3}{5} \right)

<π3\frac{\pi}{3}+ π3\frac{\pi}{3}=2π3\frac{2\pi}{3}. Hence A > B