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anonymous
 one year ago
Cosx/(1+sinx)+(1+sinx)/cos(x)=2sec(x)
I don't know how this is done at all and its so frustrating __
anonymous
 one year ago
Cosx/(1+sinx)+(1+sinx)/cos(x)=2sec(x) I don't know how this is done at all and its so frustrating __

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mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\(\dfrac{\cos x}{1 + \sin x} + \dfrac{1 + \sin x}{\cos x} = 2 \sec x\) \(\dfrac{\cos x}{1 + \sin x} \times \dfrac{\cos x}{\cos x} + \dfrac{1 + \sin x}{\cos x} \times \dfrac{1 + \sin x}{1 + \sin x}= 2 \sec x\) \(\dfrac{\cos^2 x}{(1 + \sin x) \cos x} + \dfrac{(1 + \sin x)^2}{(1 + \sin x) \cos x } = 2 \sec x\) \(\dfrac{\color{red}{\cos^2 x} + 1 + 2 \sin x + \color{red}{\sin^2 x}} {(1 + \sin x) \cos x } = 2 \sec x\) \(\dfrac{1+ 2 \sin x + \color{red}{1}} {(1 + \sin x) \cos x } = 2 \sec x\) \(\dfrac{2 + 2 \sin x}{1 + \sin x)\cos x} = 2 \sec x\) \(\dfrac{2\cancel{(1 + \sin x)}}{\cancel{(1 + \sin x)}\cos x} = 2 \sec x\) \(\dfrac{2}{\cos x} = 2 \sec x\) \(2 \sec x = 2 \sec x\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Start simply by finding the LCD of the two fractions on the left sides and adding the fractions together. Then simplify using the identity \(\sin ^2 x + \cos ^2 x = 1\) shown in red above.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathstudent55 you are soo amazing! You seriously made it look so easy lol I appreciate you taking the time to figure it out!!

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1You are very welcome. I was glad to be of help.
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