Current Affairs JEE Main & Advanced

Category : JEE Main & Advanced

(1) \[2{{\sin }^{-1}}x={{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}})\],                If \[-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}\]\[\]

(2) \[2{{\sin }^{-1}}x=\pi -{{\sin }^{-1}}2x\sqrt{1-{{x}^{2}}}\],          If \[\frac{1}{\sqrt{2}}\le x\le 1\]

(3) \[2{{\sin }^{-1}}x=-\pi -{{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}})}\],     If \[-1\le x\le \frac{-1}{\sqrt{2}}\]

(4) \[3{{\sin }^{-1}}x={{\sin }^{-1}}(3x-4{{x}^{3}}),\]                        If \[\frac{-1}{2}\le x\le \frac{1}{2}\]

(5) \[3{{\sin }^{-1}}x=\pi -{{\sin }^{-1}}(3x-4{{x}^{3}})\],                  If \[\frac{1}{2}<x\le 1\]

(6) \[3{{\sin }^{-1}}x=-\pi -{{\sin }^{-1}}(3x-4{{x}^{3}}),\]                If \[-1\le x<-\frac{1}{2}\]

(7) \[2{{\cos }^{-1}}x={{\cos }^{-1}}(2{{x}^{2}}-1)\],                         If \[0\le x\le 1\]

(8) \[2{{\cos }^{-1}}x=2\pi -{{\cos }^{-1}}(2{{x}^{2}}-1)\],                                If \[-1\le x\le 0\]

(9) \[3{{\cos }^{-1}}x={{\cos }^{-1}}(4{{x}^{3}}-3x)\],                       If \[\frac{1}{2}\le x\le 1\]

(10) \[3{{\cos }^{-1}}x=2\pi -{{\cos }^{-1}}(4{{x}^{3}}-3x),\]            If \[-\frac{1}{2}\le x\le \frac{1}{2}\]

(11)  \[3{{\cos }^{-1}}x=2\pi +{{\cos }^{-1}}(4{{x}^{3}}-3x),\]         If \[-1\le x\le -\frac{1}{2}\]

(12)  \[2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)\],                     If \[-1<x\le 1\]

(13)  \[2{{\tan }^{-1}}x=\pi +{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)\] ,                           If \[x>1\]

(14) \[2{{\tan }^{-1}}x=-\pi +{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)\],                            If \[x<-1\]

(15) \[2{{\tan }^{-1}}x={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] ,                   If \[-1\le x\le 1\]

(16) \[2{{\tan }^{-1}}x=\pi -{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] ,                            If \[x>1\]

(17) \[2{{\tan }^{-1}}x=-\pi -{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] ,                           If \[x<-1\]

(18) \[2{{\tan }^{-1}}x={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\],                                If \[0\le x<\infty \]

(19) \[2{{\tan }^{-1}}x=-{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] ,                             If \[-\infty <x\le 0\]

(20) \[3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\],                              If \[\frac{-1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\]

(21) \[3{{\tan }^{-1}}x=\pi +{{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\] ,    If \[x>\frac{1}{\sqrt{3}}\]

(22) \[3{{\tan }^{-1}}x=-\pi +{{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\] ,   If \[x<-\frac{1}{\sqrt{3}}\]

(23)  \[{{\tan }^{-1}}\left[ \frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right]={{\sin }^{-1}}\frac{x}{a}\]           

(24) \[{{\tan }^{-1}}\left[ \frac{3{{a}^{2}}x-{{x}^{3}}}{a({{a}^{2}}-3{{x}^{2}})} \right]=3{{\tan }^{-1}}\frac{x}{a}\]

(25) \[{{\tan }^{-1}}\left[ \frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]=\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]

(26) \[{{\tan }^{-1}}\sqrt{\frac{1-x}{1+x}}=\frac{1}{2}{{\cos }^{-1}}x\]