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# Sin(x)/Cos(x)

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Sin(x)/Cos(x) + Cos(x)/1 + Sin(x)

Guest Jun 12, 2015

#1
+92193
+10

I assume that you really mean this

Sin(x)/Cos(x) + Cos(x)/(1 + Sin(x) )

$$\\\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)}{1 + Sin(x) }\\\\ =\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)(1-sin(x)) }{(1 + Sin(x))(1-sin(x)) }\\\\ =\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)(1-sin(x)) }{(1-sin^2(x)) }\\\\ =\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)(1-sin(x)) }{cos^2(x) }\\\\ =\frac{1}{Cos(x)}\\\\ =sec(x)$$

Melody  Jun 12, 2015
Sort:

#1
+92193
+10

I assume that you really mean this

Sin(x)/Cos(x) + Cos(x)/(1 + Sin(x) )

$$\\\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)}{1 + Sin(x) }\\\\ =\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)(1-sin(x)) }{(1 + Sin(x))(1-sin(x)) }\\\\ =\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)(1-sin(x)) }{(1-sin^2(x)) }\\\\ =\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)(1-sin(x)) }{cos^2(x) }\\\\ =\frac{1}{Cos(x)}\\\\ =sec(x)$$

Melody  Jun 12, 2015

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