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anonymous
 one year ago
Verify the identity.
((cos x)/1+sin x) + ((1+sinx)/cos x) =2secx
anonymous
 one year ago
Verify the identity. ((cos x)/1+sin x) + ((1+sinx)/cos x) =2secx

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ cosx }{ 1+\sin x }+\frac{ 1+\sin x }{ \cos x }=2secx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know if I'm doing this right. I used LCD.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[LEFT SIDE: \frac{ \cos^2x +(1+\sin x)(1+\sin x)}{ \cos(1+sinx) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \frac{ \cos^2x +1+2sinx+sin^2x}{ \cos(1+sinx) }\]

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{(\cos x)(1\sin x)}{(1+\sin x)(1\sin x)}+\frac{(1+\sin x)(\cos x)}{\cos ^2x}=\frac{\cos x\cos x \sin x}{1\sin ^2x}+\frac{\cos x+\cos x \sin x}{\cos ^2x}\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0change cos^2 by 1sin^2x identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, am I doing it right?

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{2\cos x}{\cos ^2x}=\frac{2}{\cos x}=2\sec x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Mertsj What did you do first?

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0Multiply first fraction by (1sinx)/(1sinx)

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0Multiply second fraction by cosx/cosx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh is it like multiplying the conjugate?

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0it's really setting up cos^2x in both denominators

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0Because 1sin^2x=cos^2x

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge\rm \frac{ \color{Red}{\cos^2x} +1 +2sinx + \sin^2 }{ cosx(1+sinx) }\] \[\large\rm \frac{ \color{Red}{1sin^2x} +1 +2sinx + \sin^2 }{ cosx(1+sinx) }\] sin^2 cancel each other out you will get \[\large\rm \frac{ \color{Red}{1\cancel{sin^2x}} +1 +2sinx +\cancel{ \sin^2} }{ cosx(1+sinx) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh so what I was doing is wrong. Haha! Thanks!

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0Not wrong, such not very effective.l

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like it the solution would be longer? :D

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge\rm \frac{ 2+2sinx }{ cosx(1+sinx) }\] take out 2 \[\rm \frac{ 2(1+sinx) }{ cosx(1+sinx)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There's no "it" in my last sentence. Haha.

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0Typically there are a variety of ways to solve these identities.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhhh @Nnesha thank you so much! :) You did the whole solution for me.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0true but i think the way you were doing is easy :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's why this is harder because there are so many ways to do it and it's so frustrating haha!

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0and like mertsj said there are more than 2 ways to verify the identities
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