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Use your graphing calculator to determine if the equation appears to be an identity or not by graphing the left expression and right expression together. If so, verify the identity. If not, find a counterexample. sec x + cos x = tan x sin x

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vicobusso

Answer:

[tex](sec x)+(cos x)=(tan x)*(sin x)\\identities\\(sec x)= 1/cos x\\(tan x)=sin x/cos x\\\\and\\(cos x)^{2}+(sin x)^{2}=1\\(cos x)^{2}=1-(sin x)^{2}\\\\(1/cos x )+(cos x) = (sin x)*(sin x)/(cos x)\\ [(1/cos x)+(cos x)]*(cos x)=(sin x)^{2}\\\\1+(cos x)^{2}=(sin x)^{2}\\1+1-(sin x)^{2}=(sin x)^{2}\\2=2*(sin x)^{2}\\2/2=(sin x)^{2}\\1=(sin x)^{2}\\\\This is not true ,so\\(sec x)+(cos x)\neq (tan x)*(sin x)[/tex]

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