Jwashington471
Jwashington471
13-02-2017
Mathematics
contestada
(Secx/sinx)-(sinx/cosx)
Respuesta :
LammettHash
LammettHash
13-02-2017
[tex]\dfrac{\sec x}{\sin x}-\dfrac{\sinx }{\cos x}=\dfrac{\sec x\cos x}{\sin x\cos x}-\dfrac{\sin^2x}{\sin x\cos x}=\dfrac{1-\sin^2x}{\sin x\cos x}=\dfrac{\cos^2x}{\sin x\cos x}=\dfrac{\cos x}{\sin x}=\cot x[/tex]
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