Can someone help prove following identities

[nazar]

[MATH]1/(1+sinx)=(secx-sinxsecx)/cosx[/MATH] what I am doing is multiplying right side both numerator and denominator by 1+sinx and try to go from there but cant get 1 in numerator tried so many different ways

 

[nazar]

I think in order to solve this I need to get [MATH]sin^2x+cos^3x[/MATH] in numorator so I can divide it by cosx (1+sinx) and get 1/(1+sinx) or smth like that just tried like 20 different ways using all sorts of identities and cant get there

 

[tkhunny]

I think in order to solve this I need to get [MATH]sin^2x+cos^3x[/MATH] in numorator so I can divide it by cosx (1+sinx) and get 1/(1+sinx) or smth like that just tried like 20 different ways using all sorts of identities and cant get there

Why not multiple both denominators by 1 - sin(x)?
Maybe factor the RHS numerator.

 

[Deleted member 4993]

[MATH]1/(1+sinx)=(secx-sinxsecx)/cosx[/MATH] what I am doing is multiplying right side both numerator and denominator by 1+sinx and try to go from there but cant get 1 in numerator tried so many different ways

Start with RHS:

\(\displaystyle \displaystyle{\dfrac{sec(x) - sin(x) * sec(x)}{cos(x)}}\)

\(\displaystyle = \ \displaystyle{\dfrac{sec(x) * [1- sin(x)]}{cos(x)}}\)

\(\displaystyle = \ \displaystyle{\dfrac{ 1- sin(x)}{cos^2(x)}}\)

\(\displaystyle = \ \displaystyle{\dfrac{ 1 \ - \ sin(x)}{1 \ - \ sin^2(x)}}\)

continue......

 

[nazar]

Why not multiple both denominators by 1 - sin(x)?
Maybe factor the RHS numerator.

For me it just makes the right side much more complicated. How would u solve that?

 

[nazar]

Start with RHS:

\(\displaystyle \displaystyle{\dfrac{sec(x) - sin(x) * sec(x)}{cos(x)}}\)

\(\displaystyle = \ \displaystyle{\dfrac{sec(x) * [1- sin(x)]}{cos(x)}}\)

\(\displaystyle = \ \displaystyle{\dfrac{ 1- sin(x)}{cos^2(x)}}\)

\(\displaystyle = \ \displaystyle{\dfrac{ 1 \ - \ sin(x)}{1 \ - \ sin^2(x)}}\)

continue......

Oh okay and then multiply right side by 1-sinx which makes it 1-sinx/cos^2x think I got it now that really helps thnx

 

[nazar]

Start with RHS:

\(\displaystyle \displaystyle{\dfrac{sec(x) - sin(x) * sec(x)}{cos(x)}}\)

\(\displaystyle = \ \displaystyle{\dfrac{sec(x) * [1- sin(x)]}{cos(x)}}\)

\(\displaystyle = \ \displaystyle{\dfrac{ 1- sin(x)}{cos^2(x)}}\)

\(\displaystyle = \ \displaystyle{\dfrac{ 1 \ - \ sin(x)}{1 \ - \ sin^2(x)}}\)

continue......

But how did you get from - sinx*secx to 1-sinx?

 

[ksdhart2]

But how did you get from - sinx*secx to 1-sinx?

Let \(w = \sec(x)\) and \(y = \sin(x)\). Then:

\(\displaystyle \sec(x) - \sin(x) \sec(x) = w - yw = \text{???}\)

(Hint: Factor)

 

[nazar]

This is the stupedist chapter I had in my life. I still dont get it f....
Thought I did but I dont...

 

[Deleted member 4993]

But how did you get from - sinx*secx to 1-sinx?

Are you just staring at the screen or working with pencil-paper?

 

[nazar]

 

[Deleted member 4993]

View attachment 12359

\(\displaystyle sec(x) = \frac{1}{cos(x)}\)

so

\(\displaystyle \dfrac{sec(x)}{cos(x)} \ = \ \dfrac{\frac{1}{cos(x)}}{cos(x)} \ = \ \dfrac{1}{cos^2(x)}\)

see it now.....

 

[nazar]

Oh ok so I just have to divide secx by cosx and dont touch the 1-sinx part like how am I suppose to know that? And what am I doing wrong on the left side, thnx for the replies

 

[Otis]

... like how am I suppose to know that? ...

By practicing, with a wide variety of exercises. After sufficient exposure, things become easier.

(We learn by doing.)

?