Question

Write the all formula of inverse trigonometry.

Answer 1

Inverse trigonometric functions.
      Sin⁻¹ x = asin x = arcsin x.
      If Sin⁻¹ x = y, then x = Sin y.

$Domain, Range:\\\\Sin^{-1}x : Domain: [-1, 1].\ \ Range :[ -\pi/2, \pi/2]\\Cos^{-1}x : Domain: [-1, 1].\ \ Range :[ 0, \pi]\\Tan^{-1}x : Domain: [-\infty, \infty].\ \ \ Range :[ -\pi/2, \pi/2]\\Cot^{-1}x : Domain: [-\infty, \infty].\ \ \ Range :[ 0, \pi]\\Cosec^{-1}x : Domain: (-\infty, -1] U [1, \infty).\ \ \ Range :[ -\pi/2, 0) U (0, \pi/2]\\Sec^{-1}x : Domain: (-\infty, -1] U [1, \infty).\ \ \ Range : ( -\pi/2, \pi/2)$

$Formulas:\\\\Cosec^{-1}x=Sin^{-1}{\frac{1}{x}}.\\Sec^{-1}x= Cos^{-1}{\frac{1}{x}}\\Sin^{-1} (- x) = - Sin^{-1}x.\\Cos^{-1}(-x) = \pi - Cos^-1}x.\\Tan^{-1}(-x)= -tan^{-1}x.\\Cot^{-1}(-x)=\pi - tan^{-1}x.$

$Cot^{-1}x=Tan^{-1}{\frac{1}{x}}\\ Sin^{-1}x + COs^{-1}x=\pi/2.\\Cosec^{-1}x + sec^{-1}x = \pi/2\\\\Sin^{-1}x+sin^{-1}y=Sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})\\Cos^{-1}x + cos^{-1}y = Cos^{-1}(x y - \sqrt{(1-x^2)(1-y^2)})\\Cos^{-1}x - cos^{-1}y = Cos^{-1}(x y + \sqrt{(1-x^2)(1-y^2)})$

$Tan^{-1}x + Tan^{-1}y = Tan^{-1}(\frac{x+y}{1-xy})\\\\Sin^{-1} (\frac{2x}{1+x^2}) = 2 Tan^{-1}x = Cos^{-1} (\frac{1-x^2}{1+x^2})\\\\Tan^{-1}(\frac{2 x}{1-x^2})=2 Tan^{-1} x\\\\Tan^{-1}x + Cot^{-1}x = \pi/2\\Cot^{-1} x =Tan^{-1}{\frac{1}{x}}\\\\acos\ x=asin {\sqrt{1-x^2}}\\acos\ x=\frac{1}{2} acos(2x^2-1)\\asin\ x=\frac{1}{2} acos(1-2x^2)\\atan\ x=asin(\frac{x}{\sqrt{1+x^2}})$

Answer 2

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