Properties of Inverse Trigonometric Functions with Example

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.

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