Here, you will learn all the properties of inverse trigonometric functions class 12 with examples.
Let’s begin –
Properties of Inverse Trigonometric functions
Property – 1
(i) y = \(sin^{-1}(sinx)\), x \(\in\) R, y \(\in\) (-\(\pi\over 2\), \(\pi\over 2\)) periodic with period \(2\pi\)
and it is an odd function.(ii) y = \(cos^{-1}(cosx)\), x \(\in\) R, y \(\in\) [0, \(\pi\)], periodic with period \(2\pi\) and it is an even function.
(iii) y = \(tan^{-1}(tanx)\), x \(\in\) R – { (2n-1)\(\pi\over 2\), n \(\in\) I }, y \(\in\) (-\(\pi\over 2\), \(\pi\over 2\)) periodic with period \(\pi\) and it is an odd function.
(iv) y = \(cot^{-1}(cotx)\), x \(\in\) R – { n\(\pi\), n \(\in\) I }, y \(\in\) (0, \(\pi\)) periodic with period \(\pi\) and neither even or odd function.
(v) y = \(cosec^{-1}(cosecx)\), x \(\in\) R – { n\(\pi\), n \(\in\) I }, y \(\in\) [-\(\pi\over 2\), 0] \(\cup\) (0, \(\pi\over 2\)] periodic with period \(2\pi\) and it is an odd function.
(vi) y = \(sec^{-1}(secx)\), x \(\in\) R – { (2n-1)\(\pi\over 2\), n \(\in\) I }, y \(\in\) [0, \(\pi\over 2\)] \(\cup\) (\(\pi\over 2\), \(\pi\)], y is periodic with period \(2\pi\) and it is an even function.
Example : Evaluate \(sin^{-1}(sin10)\)
Solution : We know that \(sin^{-1}(sinx)\) = x, if \(-\pi\over 2\) \(\le\) x \(\le\) \(\pi\over 2\)
Here, x = 10 radians which does not lie between -\(\pi\over 2\) and \(\pi\over 2\)
But, \(3\pi\) – x i.e. \(3\pi\) – 10 lie between -\(\pi\over 2\) and \(\pi\over 2\)
Also, sin(\(3\pi\) – 10) = sin 10
\(\therefore\) \(sin^{-1}(sin10)\) = \(sin^{-1}(sin(3\pi – 10)\) = (\(3\pi\) – 10)
Property – 2
(i) \(sin^{-1}x\) + \(cos^{-1}x\) = \(\pi\over 2\)
(ii) \(tan^{-1}x\) + \(cot^{-1}x\) = \(\pi\over 2\)
(iii) \(cosec^{-1}x\) + \(sec^{-1}x\) = \(\pi\over 2\)
Property – 3
(i) \(sin^{-1}(-x)\) = -\(sin^{-1}x\)
(ii) \(cosec^{-1}(-x)\) = -\(cosec^{-1}x\)
(iii) \(tan^{-1}(-x)\) = -\(tan^{-1}x\)
(iv) \(cot^{-1}(-x)\) = \(\pi\) – \(cot^{-1}x\)
(v) \(cos^{-1}(-x)\) = \(\pi\) – \(cos^{-1}x\)
(vi) \(sec^{-1}(-x)\) = \(\pi\) – \(sec^{-1}x\)
Property – 4
(i) \(cosec^{-1}x\) = \(sin^{-1}{1\over x}\)
(ii) \(sec^{-1}x\) = \(cos^{-1}{1\over x}\)
(iii) \(cot^{-1}x\) = \(\begin{cases} tan^{-1}{1\over x}, & \text{if}\ x > 0 \\
\pi + tan^{-1}{1\over x}, & \text{if}\ x < 0 \end{cases}\)
Example : Find the value of x if \(cos^{-1}(-x)\) + \(tan^{-1}(-x)\) – 2\(sin^{-1}x\) + \(sec^{-1}({-1\over x})\) = \(\pi\over 4\) for |x| \(\le\) 1.
Solution : \(\pi\) – \(cos^{-1}(x)\) – \(tan^{-1}(x)\) – 2\(sin^{-1}x\) + \(cos^{-1}(-x)\) = \(\pi\over 4\)
\(\pi\) – \(cos^{-1}(x)\) – \(tan^{-1}(x)\) – 2\(sin^{-1}x\) + \(\pi\) – \(cos^{-1}(-x)\) = \(\pi\over 4\)
2\(\pi\) – 2(\(sin^{-1}x\) + \(cos^{-1}x\)) – \(\pi\over 4\) = \(tan^{-1}(x)\)
2\(\pi\) – \(\pi\) – \(\pi\over 4\) = \(tan^{-1}(x)\) \(\implies\) \(tan^{-1}(x)\) = \(3\pi\over 4\) Hence no solution.
Property – 5
(i)
(a) \(tan^{-1}x\) + \(tan^{-1}y\) = \(\begin{cases} tan^{-1}{{x+y}\over {1-xy}}, & \text{where}\ x > 0, y > 0 & xy < 1 \\ \pi + tan^{-1}{{x+y}\over {1-xy}}, & \text{where}\ x > 0, y > 0 & xy > 1 \\ {\pi\over 2} , & \text{where}\ x > 0, y > 0 & xy = 1 \end{cases}\)
(b) \(tan^{-1}x\) – \(tan^{-1}y\) = \(tan^{-1}{{x-y}\over {1+xy}}\)
(c) \(tan^{-1}x\) + \(tan^{-1}y\) + \(tan^{-1}z\) = \(tan^{-1}[{{x+y+z-xyz}\over {1-xy-yz-zx}}]\)
(ii)
(a) \(sin^{-1}x\) + \(sin^{-1}y\) = \(\begin{cases} sin^{-1}[{x\sqrt{1-y^2} + y{\sqrt{1-x^2}}}], & \text{where}\ x > 0, y > 0 & (x^2 + y^2) \le 1 \\ \pi – sin^{-1}[{x\sqrt{1-y^2} + y{\sqrt{1-x^2}}}], & \text{where}\ x > 0, y > 0 & (x^2 + y^2) > 1 \end{cases}\)
(b) \(sin^{-1}x\) – \(sin^{-1}y\) = \(sin^{-1}[{x\sqrt{1-y^2} – y{\sqrt{1-x^2}}}]\), where x > 0, y > 0
(iii)
(a) \(cos^{-1}x\) + \(cos^{-1}y\) = \(cos^{-1}[xy – {\sqrt{1-y^2}{\sqrt{1-x^2}}}]\), where x > 0, y > 0
(b) \(cos^{-1}x\) – \(cos^{-1}y\) = \(\begin{cases} cos^{-1}[xy + {\sqrt{1-y^2}{\sqrt{1-x^2}}}]; x < y, \ x, y > 0 \\ – cos^{-1}[xy + {\sqrt{1-y^2}{\sqrt{1-x^2}}}], x > y, \ x, y > 0 \end{cases}\)
Example : Prove that : \(tan^{-1}{1\over 7}\) + \(tan^{-1}{1\over 13}\) = \(tan^{-1}{2\over 9}\)
Solution : L.H.S = \(tan^{-1}{1\over 7}\) + \(tan^{-1}{1\over 13}\)
= \(tan^{-1}[{{{1\over 7}+{1\over 13}}\over {1 – {1\over 7}\times{1\over 13}}}]\) { \(\because\) \(tan^{-1}x\) + \(tan^{-1}y\) = \(tan^{-1}{{x+y}\over {1-xy}}\); if xy < 1 }
= \(tan^{-1}({20\over 90})\) = \(tan^{-1}({2\over 9})\) = R.H.S.