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It helps to convert everything to sines and cosines. Then you can often do lots of simplifications.Reminder:
sec A = 1 / cos A
tan A = sin A / cos A
cosec A = 1 / sin A
If you are unsure whether the two actually ARE equal, try evaluating both expressions (left and right) for some arbitrary value (for example, 10 degrees). If the two are NOT equal, then the expression are of course NOT equal. But if they ARE equal, you still need to prove that - since the claim is basically that the expressions are equal for ALL values of the variable.
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How do you prove the following identity sec x - cos x equals sin x tan x?
you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx
How do you put sec in a TI 84 calculator?
For any calculator Sec(Secant) = 1/Cos Csc (Cosecant) = 1/ Sin Cot (Cotangent) = 1/Tan
What does cosx divided by 1-sinx equal?
cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x
Which functions have a period of Pi?
sin(2x), cos(2x), cosec(2x), sec(2x), tan(x) and cot(x) are all possible functions.
What is the derivative of y equals sec x?
sec(x)tan(x)
