Verify the identity. ((cos x)/1+sin x) + ((1+sinx)/cos x) =2secx
\[\frac{ cosx }{ 1+\sin x }+\frac{ 1+\sin x }{ \cos x }=2secx\]
I don't know if I'm doing this right. I used LCD.
\[LEFT SIDE: \frac{ \cos^2x +(1+\sin x)(1+\sin x)}{ \cos(1+sinx) }\]
\[ \frac{ \cos^2x +1+2sinx+sin^2x}{ \cos(1+sinx) }\]
\[\frac{(\cos x)(1-\sin x)}{(1+\sin x)(1-\sin x)}+\frac{(1+\sin x)(\cos x)}{\cos ^2x}=\frac{\cos x-\cos x \sin x}{1-\sin ^2x}+\frac{\cos x+\cos x \sin x}{\cos ^2x}\]
change cos^2 by 1-sin^2x identity
Wait, am I doing it right?
\[\frac{2\cos x}{\cos ^2x}=\frac{2}{\cos x}=2\sec x\]
@Mertsj What did you do first?
Multiply first fraction by (1-sinx)/(1-sinx)
Multiply second fraction by cosx/cosx
Ohhh is it like multiplying the conjugate?
it's really setting up cos^2x in both denominators
Because 1-sin^2x=cos^2x
\[\huge\rm \frac{ \color{Red}{\cos^2x} +1 +2sinx + \sin^2 }{ cosx(1+sinx) }\] \[\large\rm \frac{ \color{Red}{1-sin^2x} +1 +2sinx + \sin^2 }{ cosx(1+sinx) }\] sin^2 cancel each other out you will get \[\large\rm \frac{ \color{Red}{1\cancel{-sin^2x}} +1 +2sinx +\cancel{ \sin^2} }{ cosx(1+sinx) }\]
Ohhh so what I was doing is wrong. Haha! Thanks!
i don't think so..
Not wrong, such not very effective.l
Like it the solution would be longer? :D
\[\huge\rm \frac{ 2+2sinx }{ cosx(1+sinx) }\] take out 2 \[\rm \frac{ 2(1+sinx) }{ cosx(1+sinx)}\]
There's no "it" in my last sentence. Haha.
Typically there are a variety of ways to solve these identities.
Ohhhh @Nnesha thank you so much! :) You did the whole solution for me.
true but i think the way you were doing is easy :D
That's why this is harder because there are so many ways to do it and it's so frustrating haha!
and like mertsj said there are more than 2 ways to verify the identities
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