Simplify the expression. tan(sin^-1(x))?
Can someone help me simplify the expression (\tan(\sin^{-1}(x)))? I would appreciate a detailed step-by-step explanation for solving this problem.
4 Answers
tan [sin^-1 (x)]let u = sin^-1 (x)so, sin u = x now u draw a right-angled triangle with an angle u :from sin u = x , the opp. is x and the hypotenus is 1, then u find the adj. which is √(1 – x^2)tan [sin^-1 (x)] = tan u = opp./adj.= x / √(1 – x^2)therefore tan [sin^-1 (x)] = x / √(1 – x^2)
If by sin^-1(x) you mean the inverse function of sin then proceed otherwise disregard this this answer. OK so think that sin^-1(x) = T then by applying the sin function to both sides we have sin(sin^-1(x)) = sin(T)x = sin(T) so, sin(T) = xnow since sin(T) = opposite/hypotenuse = xso let the hypotenuse = 1 so the opposite side of T = xso we need to find the measurement of the side adjacent to T using the Pythagorean theorem we get the square root of (1 – x^2)now since tan = opposite/adjacentwe havetan(sin^-1(x)) = tan(T) = x / [root(1-x^2) ]
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