Find sec x tan x=-5 cos x >0

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gsenviro | College Teacher | (Level 1) Educator Emeritus

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There is some confusion, I think your question is:

sec x tan x = -5 , given that cos x > 0 

Using the above mentioned equality, and sec x  = 1/cosx and tan x  = sin x/cosx

we get, sec x tan x = (1/cos x) (sinx /cosx) = sinx/cos^2x = -5

using the identity, cos^2 x = 1-sin^x, we get

sinx/(1-sin^2x) = -5 

substituting sinx = a, we get

a/(1-a^2) = -5 

or a = -5 + 5 a^2 

or 5 a^2 -a -5 = 0

Solving this quadratic equation for a,

a = [-(-1) +- sqrt (1^2 -4(-5)(5))] / (2x5) .

we get 2 values of a, -1.105 and 0.905

since sin x cannot be more than 1, the only valid solution is 0.905

thus, sin x  = 0.905

or x = 64.82 degrees (or 0.36 pi radians)

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