• # question_answer The greater of the two angles $A=2{{\tan }^{-1}}(2\sqrt{2}-1)$and $B=3{{\sin }^{-1}}\left( \frac{1}{3} \right)+$${{\sin }^{-1}}\left( \frac{3}{5} \right)$ is A)  B                                            B)  A C)  C                                            D)  None of these

Idea $\because$$x=\sin \theta ,\theta ={{\sin }^{-1}}x,-1\le x\le 1$and $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2},x=\tan \theta ,\theta ={{\tan }^{-1}}x,-\infty <x<\infty$and$\frac{-\pi }{2}<\theta <\frac{\pi }{2}$ Given angles are A = 2 ${{\tan }^{-1}}(2\sqrt{2}-1)$ and    $B=3{{\sin }^{-1}}\left( \frac{1}{3} \right)+{{\sin }^{-1}}\left( \frac{3}{5} \right)$ Consider $A=2{{\tan }^{-1}}(2\sqrt{2}-1)=2{{\tan }^{-1}}(1.828)$ $\therefore$$A>2{{\tan }^{-1}}\sqrt{3}$       $(\because \sqrt{3}=1.732<1.828)$ $\Rightarrow$               $A>\frac{2\pi }{3}$                                       ?(i) Also       ${{\sin }^{-1}}\left( \frac{1}{3} \right)<{{\sin }^{-1}}\left( \frac{1}{2} \right)=\frac{\pi }{6}$ $\Rightarrow$               $3{{\sin }^{-1}}\left( \frac{1}{3} \right)<\frac{\pi }{6}$ So, using the identity $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta ,$we have ${{\sin }^{-1}}\left( \frac{1}{3} \right)={{\sin }^{-1}}\left( 3\times \frac{1}{3}-4{{\left( \frac{1}{3} \right)}^{3}} \right)={{\sin }^{-1}}\left( \frac{23}{27} \right)$ $={{\sin }^{-1}}(0.852)$ $\therefore$$3{{\sin }^{-1}}\left( \frac{1}{3} \right)<{{\sin }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\left\{ \because \frac{\sqrt{3}}{2}=0.868>0.852 \right\}$i.e., $3{{\sin }^{-1}}\left( \frac{1}{3} \right)<\frac{\pi }{3}$ also${{\sin }^{-1}}\left( \frac{3}{5} \right)={{\sin }^{-1}}(0.6)<{{\sin }^{-1}}\left( \frac{\sqrt{3}}{2} \right)=\frac{\pi }{3}$ $\therefore$  ${{\sin }^{-1}}\left( \frac{3}{5} \right)=\frac{\pi }{3}$ $\therefore$  $B=3{{\sin }^{-1}}\left( \frac{3}{5} \right)+\sin \left( \frac{3}{5} \right)$ $<\frac{\pi }{3}+\frac{\pi }{3}=\frac{2\pi }{3}$                  $\therefore$$B<\frac{2\pi }{3}$ Thus, from Eqs. (i) and (iii), we have A > B $\therefore$ A is the greater angle. TEST Edge Application of inverse trigonometric function based questions are asked. To solve such type of question, students are advised to understand the concept of these functions.