Current Affairs JEE Main & Advanced

Category : JEE Main & Advanced

(1) \[{{\sin }^{-1}}(\sin \theta )=\theta \],   Provided that \[-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}\],   

\[{{\cos }^{-1}}(\cos \theta )=\theta \],   Provided that \[0\le \theta \le \pi \]

\[{{\tan }^{-1}}(\tan \theta )=\theta \],   Provided that \[-\frac{\pi }{2}<\theta <\frac{\pi }{2}\],            

\[{{\cot }^{-1}}(\cot \theta )=\theta \],   Provided that \[0<\theta <\pi \]

\[{{\sec }^{-1}}(\sec \theta )=\theta \],  Provided that \[0\le \theta <\frac{\pi }{2}\] or \[\frac{\pi }{2}<\theta \le \pi \]         

\[\text{cose}{{\text{c}}^{-1}}(\text{cosec}\theta \text{)}=\theta \text{,}\]Provided that \[-\frac{\pi }{2}\le \theta <0\]or \[0<\theta \le \frac{\pi }{2}\]

(2) \[\sin ({{\sin }^{-1}}x)=x,\] Provided that \[-1\le x\le 1\],       

\[\cos ({{\cos }^{-1}}x)=x,\] Provided that \[-1\le x\le 1\]

tan \[({{\tan }^{-1}}x)=x,\] Provided that \[-\infty <x<\infty \]

\[\cot ({{\cot }^{-1}}x)=x,\] Provided that \[-\infty <x<\infty \]

\[\sec ({{\sec }^{-1}}x)=x,\] Provided that \[-\infty <x\le -1\] or \[1\le x<\infty \]

\[\text{cosec }(\text{cose}{{\text{c}}^{\text{--1}}}x)=x,\]Provided that \[-\infty <x\le -1\] or \[1\le x<\infty \]

(3) \[{{\sin }^{-1}}(-x)=-{{\sin }^{-1}}x\],     \[{{\cos }^{-1}}(-x)=\pi -{{\cos }^{-1}}x\]

\[{{\tan }^{-1}}(-x)=-{{\tan }^{-1}}x\] ,    \[{{\cot }^{-1}}(-x)=\pi -{{\cot }^{-1}}x\]      

\[{{\sec }^{-1}}(-x)=\pi -{{\sec }^{-1}}x\],   \[\text{cose}{{\text{c}}^{-1}}(-x)=-\text{cose}{{\text{c}}^{\text{--1}}}x\]

(4) \[{{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2}\],    for all \[x\in [-1,\,1]\]

\[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2}\],    for all \[x\in R\]

\[{{\sec }^{-1}}x+\text{cose}{{\text{c}}^{\text{-1}}}x=\frac{\pi }{2}\],    for all \[x\in (-\infty ,\,-1]\cup [1,\,\infty )\]

(5) Principal values for inverse circular functions

(6) Conversion property : Let,  \[{{\sin }^{-1}}x=y\] Þ \[x=\sin y\]

Þ  \[\text{cosec }y=\left( \frac{1}{x} \right)\]  Þ  \[y=\text{cose}{{\text{c}}^{\text{--1}}}\left( \frac{1}{x} \right)\]

\[{{\sin }^{-1}}x={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}={{\tan }^{-1}}\frac{x}{\sqrt{1-{{x}^{2}}}}\]

\[={{\cot }^{-1}}\frac{\sqrt{1-{{x}^{2}}}}{x}={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{x}^{2}}}} \right)=\text{cose}{{\text{c}}^{\text{--1}}}\left( \frac{1}{x} \right)\]

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

\[={{\sec }^{-1}}\frac{1}{x}=\text{cose}{{\text{c}}^{\text{--1}}}\left( \frac{1}{\sqrt{1-{{x}^{2}}}} \right)={{\cot }^{-1}}\left( \frac{x}{\sqrt{1-{{x}^{2}}}} \right)\]

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

\[={{\cot }^{-1}}\left( \frac{1}{x} \right)={{\sec }^{-1}}\sqrt{1+{{x}^{2}}}=\text{cose}{{\text{c}}^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}}{x} \right)\]

\[{{\sin }^{-1}}\left( \frac{1}{x} \right)=\text{cose}{{\text{c}}^{-1}}x\], for all \[x\in (-\infty ,1]\cup [1,\,\infty )\]

\[{{\cos }^{-1}}\left( \frac{1}{x} \right)\,={{\sec }^{-1}}x,\] for all \[x\in (-\infty ,\,1]\,\cup \,[1,\,\infty )\]

\[{{\tan }^{-1}}\left( \frac{1}{x} \right)=\left\{ \begin{matrix} {{\cot }^{-1}}x,\,\,\,\,\,\,\,\,\,\,\,\text{for }x>0  \\-\pi +{{\cot }^{-1}}x,\,\text{for }x\end{matrix} \right.\] 

(7) General values of inverse circular functions: We know that if a is the smallest angle whose sine is x, then all the angles whose sine is x can be written as \[nx+{{(-1)}^{n}}\alpha ,\] where \[n=0,\,1,\,2,.....\]. Therefore, the general value of \[{{\sin }^{-1}}x\] can be taken as \[n\pi +{{(-1)}^{n}}\alpha \]. The general value of \[{{\sin }^{-1}}x\] is denoted by \[{{\sin }^{-1}}x\].

Thus, we have 

\[si{{n}^{-1}}x=n\pi +{{(-1)}^{n}}\alpha ,\,-1\le x\le 1,\,if sin\,\alpha =x and -\frac{\pi }{2}\le \alpha \le \frac{\pi }{2}\]

Similarly, general values of other inverse circular functions are given as follows: