Inverse Trigonometric Functions - Revision Notes

 CBSE Class 12 Mathematics

Chapter-02
Inverse Trigonometric Functions

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• The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table:

Functions	Domain	Range (Principal Value Branches)
	[-1, 1]
	[-1, 1]
	R- [-1, 1]	- {0}
	R-[-1, 1]
	R
	R

•  should not be confused with . In fact  And similarly for other trigonometric functions.
• The value of an inverse trigonometric functions which lies in its principal value branch is called the principal value of that inverse trigonometric functions.
• For suitable values of domain, we have

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cot−11x=tan−1x${\cot }^{-1}\frac{1}{x}={\tan }^{-1}x$

cosec−11x=sin−1x$\cos e{c}^{-1}\frac{1}{x}={\sin }^{-1}x$

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sec−11x=cos−1x${\sec }^{-1}\frac{1}{x}={\cos }^{-1}x$

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sin−1x+sin−1y=sin−1(x1−y2−−−−−√+y1−x2−−−−−√)${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)$

cos−1x+ocos−1y=cos−1(xy−1−x2−−−−−√1−y2−−−−−√)${\cos }^{-1}x+o{\cos }^{-1}y={\cos }^{-1}\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)$

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tan−1x+tan−1y+tan−1z${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z$ = tan−1(x+y+z−xyz1−xy−yz−zx)${\tan }^{-1}\left(\frac{x+y+z-xyz}{1-xy-yz-zx}\right)$

2sin−1x=sin−1(2x1−x2−−−−−√)$2{\sin }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

2cos−1x=cos−1(2x2−1)$2{\cos }^{-1}x={\cos }^{-1}\left(2x^2-1\right)$

• 2tan−1x=sin−1(2x1+x2)$2{\tan }^{-1}x={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ = cos−1(1−x21+x2)=tan−1(2x1−x√)${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)={\tan }^{-1}\left(\frac{2x}{1-\sqrt{x}}\right)$

3sin−1x=sin−1(3x−4x3)$3{\sin }^{-1}x={\sin }^{-1}\left(3x-4x^3\right)$

3cos−1x=cos−1(4x3−3x)$3{\cos }^{-1}x={\cos }^{-1}\left(4x^3-3x\right)$

3tan−1x=tan−1(3x−x31−3x2)$3{\tan }^{-1}x={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$

Conversion:

• sin−1x=cos−11−x2−−−−−√${\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}$=tan−1x1−x2√=cot−11−x2√x${\tan }^{-1}\frac{x}{\sqrt{1-x^2}}={\cot }^{-1}\frac{\sqrt{1-x^2}}{x}$ = sec−111−x2√=cosec−11x${\sec }^{{}^{-1}}\frac{1}{\sqrt{1-x^2}}=\cos e{c}^{-1}\frac{1}{x}$
• cos−1x=sin−11−x2−−−−−√${\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}$=tan−11−x2√x=cot−1x1−x2√${\tan }^{{}^{-1}}\frac{\sqrt{1-x^2}}{x}={\cot }^{-1}\frac{x}{\sqrt{1-x^2}}$ = sec−11x=cosec−111−x2√${\sec }^{{}^{-1}}\frac{1}{x}=\cos e{c}^{-1}\frac{1}{\sqrt{1-x^2}}$
• tan−1x=sin−1x1+x2√${\tan }^{{}^{-1}}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}}$=cos−111+x2√=sec−11+x2−−−−−√${\cos }^{{}^{-1}}\frac{1}{\sqrt{1+x^2}}={\sec }^{-1}\sqrt{1+x^2}$ = cosec−11+x2√x=cot−11x$\cos e{c}^{{}^{-1}}\frac{\sqrt{1+x^2}}{x}={\cot }^{-1}\frac{1}{x}$
• cot−1x=sin−111+x2√${\cot }^{{}^{-1}}x={\sin }^{-1}\frac{1}{\sqrt{1+x^2}}$=cos−1x1+x2√=sec−11x${\cos }^{{}^{-1}}\frac{x}{\sqrt{1+x^2}}={\sec }^{-1}\frac{1}{x}$= sec−11+x2√x=cosec−11+x2−−−−−√${\sec }^{{}^{-1}}\frac{\sqrt{1+x^2}}{x}=\cos e{c}^{-1}\sqrt{1+x^2}$
• sec−1x=tan−1x2−1√1${\sec }^{{}^{-1}}x={\tan }^{-1}\frac{\sqrt{x^2-1}}{1}$=cot−11x2−1√=sin−1x2−1√x${\cot }^{{}^{-1}}\frac{1}{\sqrt{x^2-1}}={\sin }^{-1}\frac{\sqrt{x^2-1}}{x}$ = cos−11x=cosec−1xx2−1√${\cos }^{{}^{-1}}\frac{1}{x}=\cos e{c}^{-1}\frac{x}{\sqrt{x^2-1}}$
• cosec−1x=sin−11x$\cos e{c}^{{}^{-1}}x={\sin }^{-1}\frac{1}{x}$=tan−11x2−1√=cot−1x2−1−−−−−√${\tan }^{{}^{-1}}\frac{1}{\sqrt{x^2-1}}={\cot }^{-1}\sqrt{x^2-1}$=sec−1xx2−1√=cos−1x2−1√x${\sec }^{{}^{-1}}\frac{x}{\sqrt{x^2-1}}={\cos }^{-1}\frac{\sqrt{x^2-1}}{x}$

Some other properties of Inverse Trigonometric Function:

• tan−1xa2−x2√=sin−1xa${\tan }^{{}^{-1}}\frac{x}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}$
• cot−1xa2−x2√=cos−1xa${\cot }^{{}^{-1}}\frac{x}{\sqrt{a^2-x^2}}={\cos }^{-1}\frac{x}{a}$
• tan−1ax2−a2√=cosec−1xa${\tan }^{{}^{-1}}\frac{a}{\sqrt{x^2-a^2}}=\cos e{c}^{-1}\frac{x}{a}$
• cot−1ax2−a2√=sec−1xa