{"id":3851,"date":"2021-08-09T13:20:07","date_gmt":"2021-08-09T13:20:07","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3851"},"modified":"2021-11-26T16:28:23","modified_gmt":"2021-11-26T10:58:23","slug":"properties-of-inverse-trigonometric-functions","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/properties-of-inverse-trigonometric-functions\/","title":{"rendered":"Properties of Inverse Trigonometric Functions with Example"},"content":{"rendered":"
Here, you will learn all the properties of inverse trigonometric functions class 12 with examples.<\/p>\n
Let’s begin –<\/p>\n
Property – 1<\/strong><\/p>\n (i)\u00a0 y = \\(sin^{-1}(sinx)\\), x \\(\\in\\) R, y \\(\\in\\) (-\\(\\pi\\over 2\\), \\(\\pi\\over 2\\)) periodic with period \\(2\\pi\\) (ii)\u00a0 y = \\(cos^{-1}(cosx)\\), x \\(\\in\\) R, y \\(\\in\\) [0, \\(\\pi\\)], periodic with period \\(2\\pi\\) and it is an even function.<\/p>\n (iii)\u00a0 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.<\/p>\n (iv)\u00a0 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.<\/p>\n (v)\u00a0 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.<\/p>\n (vi)\u00a0 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.<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Evaluate \\(sin^{-1}(sin10)\\)<\/p>\n Solution : <\/span>We know that \\(sin^{-1}(sinx)\\) = x, if \\(-\\pi\\over 2\\) \\(\\le\\) x \\(\\le\\) \\(\\pi\\over 2\\) Property – 2<\/strong><\/p>\n (i) \\(sin^{-1}x\\) + \\(cos^{-1}x\\) = \\(\\pi\\over 2\\)<\/p>\n (ii) \\(tan^{-1}x\\) + \\(cot^{-1}x\\) = \\(\\pi\\over 2\\)<\/p>\n (iii) \\(cosec^{-1}x\\) + \\(sec^{-1}x\\) = \\(\\pi\\over 2\\)<\/p>\n<\/blockquote>\n Property – 3<\/strong><\/p>\n (i) \\(sin^{-1}(-x)\\) = -\\(sin^{-1}x\\)<\/p>\n (ii) \\(cosec^{-1}(-x)\\) = -\\(cosec^{-1}x\\)<\/p>\n (iii) \\(tan^{-1}(-x)\\) = -\\(tan^{-1}x\\)<\/p>\n (iv) \\(cot^{-1}(-x)\\) = \\(\\pi\\) – \\(cot^{-1}x\\)<\/p>\n (v) \\(cos^{-1}(-x)\\) = \\(\\pi\\) – \\(cos^{-1}x\\)<\/p>\n (vi) \\(sec^{-1}(-x)\\) = \\(\\pi\\) – \\(sec^{-1}x\\)<\/p>\n<\/blockquote>\n Property – 4<\/strong><\/p>\n (i) \\(cosec^{-1}x\\) = \\(sin^{-1}{1\\over x}\\)<\/p>\n (ii) \\(sec^{-1}x\\) = \\(cos^{-1}{1\\over x}\\)<\/p>\n (iii) \\(cot^{-1}x\\) = \\(\\begin{cases} tan^{-1}{1\\over x}, & \\text{if}\\ x > 0 \\\\ Example : <\/span> 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.<\/p>\n Solution : <\/span>\\(\\pi\\) – \\(cos^{-1}(x)\\) – \\(tan^{-1}(x)\\) – 2\\(sin^{-1}x\\) + \\(cos^{-1}(-x)\\) = \\(\\pi\\over 4\\) Property – 5<\/strong><\/p>\n (i) <\/p>\n (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}\\)<\/p>\n (b) \\(tan^{-1}x\\) – \\(tan^{-1}y\\) = \\(tan^{-1}{{x-y}\\over {1+xy}}\\)<\/p>\n (c) \\(tan^{-1}x\\) + \\(tan^{-1}y\\) + \\(tan^{-1}z\\) = \\(tan^{-1}[{{x+y+z-xyz}\\over {1-xy-yz-zx}}]\\)<\/p>\n (ii)<\/p>\n (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}\\)<\/p>\n (b) \\(sin^{-1}x\\) – \\(sin^{-1}y\\) = \\(sin^{-1}[{x\\sqrt{1-y^2} – y{\\sqrt{1-x^2}}}]\\), where x > 0, y > 0<\/p>\n (iii)<\/p>\n (a) \\(cos^{-1}x\\) + \\(cos^{-1}y\\) = \\(cos^{-1}[xy – {\\sqrt{1-y^2}{\\sqrt{1-x^2}}}]\\), where x > 0, y > 0<\/p>\n (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}\\)<\/p>\n<\/blockquote>\n\n\n Example : <\/span>Prove that : \\(tan^{-1}{1\\over 7}\\) + \\(tan^{-1}{1\\over 13}\\) = \\(tan^{-1}{2\\over 9}\\)<\/p>\n Solution : <\/span>L.H.S = \\(tan^{-1}{1\\over 7}\\) + \\(tan^{-1}{1\\over 13}\\)\n
and it is an odd function.<\/p>\n
\n Here, x = 10 radians which does not lie between -\\(\\pi\\over 2\\) and \\(\\pi\\over 2\\)
\n But, \\(3\\pi\\) – x i.e. \\(3\\pi\\) – 10 lie between -\\(\\pi\\over 2\\) and \\(\\pi\\over 2\\)
\n Also, sin(\\(3\\pi\\) – 10) = sin 10
\n \\(\\therefore\\) \\(sin^{-1}(sin10)\\) = \\(sin^{-1}(sin(3\\pi – 10)\\) = (\\(3\\pi\\) – 10)
<\/p>\n\n\n\n
\n
\n
\\pi + tan^{-1}{1\\over x}, & \\text{if}\\ x < 0 \\end{cases}\\)<\/p>\n<\/blockquote>\n\n\n
\n \\(\\pi\\) – \\(cos^{-1}(x)\\) – \\(tan^{-1}(x)\\) – 2\\(sin^{-1}x\\) + \\(\\pi\\) – \\(cos^{-1}(-x)\\) = \\(\\pi\\over 4\\)
\n 2\\(\\pi\\) – 2(\\(sin^{-1}x\\) + \\(cos^{-1}x\\)) – \\(\\pi\\over 4\\) = \\(tan^{-1}(x)\\)
\n 2\\(\\pi\\) – \\(\\pi\\) – \\(\\pi\\over 4\\) = \\(tan^{-1}(x)\\) \\(\\implies\\) \\(tan^{-1}(x)\\) = \\(3\\pi\\over 4\\) Hence no solution.
\n <\/p>\n\n\n\n
\n = \\(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 }
\n = \\(tan^{-1}({20\\over 90})\\) = \\(tan^{-1}({2\\over 9})\\) = R.H.S.
<\/p>\n\n\n\n